紫薯 第四章习题enmmmm
习题4-1 象棋uva1589
题意，输入帅 马 车 炮 将 的棋盘位置，保证局面合法且红方已经将军判断红方是否能将黑方将死
做法 强行一波 模拟 ，把 将 和 车 看成一样的， 10*9的棋盘而已，两个字符数组 一个是旗子位置 另一个是红方可以走的位置
黑方的 将 如果上下左右 能走的地方都是被标记过的（红方可以走的） 则GG
有个 需要注意的地方 就是炮 和 车 在列举可能情况的时候，先赋值 再判断是否循环结束
比如说车隔着前面四个格子是马 马位置也要赋值 因为可能 对面的将下一步把马给吃掉了 //因为这个一直WA 心塞
#include <cstdio>//是否能将死 对面的shuai #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; char kaiju[11][11],maybe[11][11],ch; int n,x,y,x2,y2,a,t1,t2; void RHG(int x1,int y1) { for(int i = x1-1; i > 0; i--) { maybe[i][y1] = 'B';//因为可能旗子会被帅吃掉 if(kaiju[i][y1] != 'A' && kaiju[i][y1] != 'S') break; } for(int i = x1+1; i < 4; i++) { maybe[i][y1] = 'B'; if(kaiju[i][y1] != 'A' && kaiju[i][y1] != 'S') break; } for(int i = y1+1; i < 7; i++) { maybe[x1][i] = 'B'; if(kaiju[x1][i] != 'A' && kaiju[x1][i] != 'S') break; } for(int i = y1-1; i > 3; i--) { maybe[x1][i] = 'B'; if(kaiju[x1][i] != 'A' && kaiju[x1][i] != 'S') break; } return; } void RH(int x1,int y1) { if (kaiju[x1-1][y1]=='A' && x1-2>0 && y1+1<7) maybe[x1-2][y1+1] = 'B'; if (kaiju[x1-1][y1]=='A' && x1-2>0 && y1-1>3) maybe[x1-2][y1-1] = 'B'; if (kaiju[x1+1][y1]=='A' && x1+2<4 && y1+1<7) maybe[x1+2][y1+1] = 'B'; if (kaiju[x1+1][y1]=='A' && x1+2<4 && y1-1>3) maybe[x1+2][y1-1] = 'B'; if (kaiju[x1][y1-1]=='A' && y1-2>3 && x1-1>0) maybe[x1-1][y1-2] = 'B'; if (kaiju[x1][y1-1]=='A' && y1-2>3 && x1+1<4) maybe[x1+1][y1-2] = 'B'; if (kaiju[x1][y1+1]=='A' && y1+2<7 && x1-1>0) maybe[x1-1][y1+2] = 'B'; if (kaiju[x1][y1+1]=='A' && y1+2<7 && x1+1<4) maybe[x1+1][y1+2] = 'B'; } void RC(int x1,int y1) { if (x1==1 && y1<7 && y1>3 && kaiju[2][y1]!='S' && kaiju[2][y1]!='A') maybe[3][y1] = 'B'; int j; for (j = x1-1; j > 0; j--) { if (kaiju[j][y1] == 'S' ) return; if (kaiju[j][y1] != 'A') break; }for (int i = j-1; i > 0; i--) { maybe[i][y1] = 'B'; if (kaiju[i][y1] != 'A' && kaiju[i][y1] != 'S') break; } for (j = y1-1; j >3; j--) { if (kaiju[x1][j] == 'S') return; if (kaiju[x1][j] != 'A') break; }for (int i = j-1; i > 3; i--) { maybe[x1][i] = 'B'; if (kaiju[x1][i] == 'A' && kaiju[x1][i] != 'S') break; } for( j = y1+1; j < 7; j++) { if (kaiju[x1][j] == 'S') return; if (kaiju[x1][j] != 'A') break; }for (int i = j+1; i < 7; i++) { maybe[x1][i] = 'B'; if (kaiju[x1][i] == 'A' && kaiju[x1][i]) break; } } int jiang() {//s[x][y]== G if (x-1>0 && maybe[x-1][y]=='A')return 0; if (x+1<4 && maybe[x+1][y]=='A')return 0; if (y-1>3 && maybe[x][y-1]=='A')return 0; if (y+1<7 && maybe[x][y+1]=='A')return 0; return 1; } int main() { while (scanf("%d%d%d",&n,&x,&y) == 3 && n) { int v[8][3]; memset(maybe,'A',sizeof(maybe)); memset(kaiju,'A',sizeof(kaiju)); memset(v,0,sizeof(v)); kaiju[x][y] = 'S'; for (int i = 0; i < n; i++) { char s[33]; scanf("%s%d%d", s, &x2, &y2); //printf("%s\n", s); ch = s[0]; v[i][0] = x2;v[i][1] = y2; kaiju[x2][y2] = ch; getchar(); } for (int i = 0; i < n; i++) { int x1 = v[i][0],y1 = v[i][1]; if (kaiju[x1][y1] == 'G' || kaiju[x1][y1] =='R') RHG(x1,y1); if (kaiju[x1][y1] == 'H') RH(x1,y1); if (kaiju[x1][y1] == 'C') RC(x1,y1); } puts(jiang()?"YES":"NO"); //if (jiang()) printf("YES\n"); //GG 无路可退 //else printf("NO\n"); } return 0; }
看到别人有个玄学的做法 没看懂2333 而且很短
#include <cstdio> #define RD(X) if(s[X].x||s[X].y)s[X+1].x=x,s[X+1].y=y;else s[X].x=x,s[X].y=y #define CHECKR(X) (check(s[X],m,n)&&!between(s[X],m,n)) #define CHECKC(X) (check(s[X],m,n)&&between(s[X],m,n)==1) char c; int x,y,N; struct point{int x,y;}; point s[8];//B,G,R1,R2,H1,H2,C1,C2 int mx[4]={1,-1,0,0},my[4]={0,0,1,-1}; int abs(int tt){return ((tt<0)?-tt:tt);} int mid(int a1,int a2,int a3){return ((a1<a2&&a2<a3)||(a1>a2&&a2>a3));} int check(point TT,int x2,int y2){return ((TT.x==x2&&TT.y!=y2)||(TT.y==y2&&TT.x!=x2));} int between(point T,int x2,int y2){ int xx=0; for(int i=1;i<8;i++) xx+=(mid(T.x,s[i].x,x2)&&check(s[i],x2,y2))+(mid(T.y,s[i].y,y2)&&check(s[i],x2,y2)); return xx; } int checkH(point _T,int mm,int nn){ int dm=_T.x-mm,dn=_T.y-nn,am=abs(dm),an=abs(dn); return (am+an==3&&am&&an&&!between(_T,_T.x-(int(dm/2)<<1),_T.y-(int(dn/2)<<1))); } int Try(int m,int n){ if(m<1||m>3||n<4||n>6||\ (n==s[1].y&&!between(s[1],m,n))||\ CHECKR(2)||CHECKR(3)||\ CHECKC(6)||CHECKC(7))return 0; return !(checkH(s[4],m,n)||checkH(s[5],m,n)); } int main(){ while(~scanf("%d%d%d",&N,&s[0].x,&s[0].y)&&(N||s[0].x||s[0].y)){ for(int i=1;i<8;i++)s[i].x=s[i].y=0; while(N--){ scanf(" %c%d%d",&c,&x,&y); if(c=='G')s[1].x=x,s[1].y=y; if(c=='R')RD(2); if(c=='H')RD(4); if(c=='C')RD(6); } for(int i=0;i<4;i++) if(Try(s[0].x+mx[i],s[0].y+my[i])){printf("NO\n");goto lable;} printf("YES\n");lable:; } return 0; }
习题4-2 正方形 uva201
题意 输入n行n列的小黑点，输入m条边，问能组成的正方形个数
解法：遍历所有点，先判断右 下边有无边相连，再右下 右下左 有无边 （从 1-> 9）,always WA ，不知为何
#include<algorithm> #include<cstring> #include<cstdio> #include<cmath> int n,m,v[20][20],h[20][20],a[11]; using namespace std; void panduan(int x,int y) { if(v[x][y] && h[x][y]) // { //printf(" a[1] pandusan\n"); if(v[x][y+1] && h[x+1][y]) a[1]++; if(v[x+1][y] && h[x][y+1]) { if(v[x][y+2] && v[x+1][y+2] && h[x+2][y] && h[x+2][y+1]) a[2]++; if(v[x+2][y] && h[x][y+2]) { //printf(" a[3] pandusan\n"); if(v[x][y+3] && v[x+1][y+3] && v[x+2][y+3] && h[x+3][y] && h[x+3][y+1] && h[x+3][y+2]) a[3]++; if(v[x+3][y] && h[x][y+3]) { // printf(" a[4] pandusan\n"); if(v[x][y+4] && v[x+1][y+4] && v[x+2][y+4] && v[x+3][y+4] && h[x+4][y] && v[x+4][y+1] && v[x+4][y+2] && v[x+4][y+3]) a[4]++; if(v[x+4][y] && h[x][y+4]) { // printf(" a[5] pandusan\n"); if(v[x][y+5] && v[x+1][y+5] && v[x+2][y+5] && v[x+3][y+5] && v[x+4][y+5] && h[x+5][y] && h[x+5][y+1] && h[x+5][y+2] && h[x+5][y+3] && h[x+5][y+4]) a[5]++; if(v[x+5][y] && h[x][y+5]) { if(v[x][y+6] && v[x+1][y+6] && v[x+2][y+6] && v[x+3][y+6] && v[x+4][y+6] && v[x+5][y+6] && h[x+6][y] && h[x+6][y+1] && h[x+6][y+2] && h[x+6][y+3] && h[x+6][y+4] && h[x+6][y+5]) a[6]++; if(v[x+6][y] && h[x][y+6]) { if(v[x][y+7] && v[x+1][y+7] && v[x+2][y+7] && v[x+3][y+7] && v[x+4][y+7] && v[x+5][y+7] && v[x+6][y+7] && h[x+7][y] && h[x+7][y+1] && h[x+7][y+2] && h[x+7][y+3] && h[x+7][y+4] && h[x+7][y+5] && h[x+7][y+6]) a[7]++; if(v[x+7][y] && h[x][y+7]) { if(v[x][y+8] && v[x+1][y+8] && v[x+2][y+8] && v[x+3][y+8] && v[x+8][y+8] && v[x+5][y+8] && v[x+6][y+8] && v[x+7][y+8] && h[x+8][y] && h[x+8][y+1] && h[x+8][y+2] && h[x+8][y+3] && h[x+8][y+4] && h[x+8][y+5] && h[x+8][y+6] && h[x+8][y+7]) a[8]++; if(v[x+8][y] && h[x][y+8]) { if(v[x][y+9] && v[x+1][y+9] && v[x+2][y+9] && v[x+3][y+9] && v[x+4][y+9] && v[x+5][y+9] && v[x+6][y+9] && v[x+7][y+9] && v[x+8][y+9] && h[x+9][y] && h[x+9][y+1] && h[x+9][y+2] && h[x+9][y+3] && h[x+9][y+4] && h[x+9][y+5] && h[x+9][y+6] && h[x+9][y+7] && h[x+9][y+8]) a[9]++; return; } else return; } else return; } else return; } else return; } else return; } else return; } else return; } else return; } else return; } // Problem #1 // 2 square (s) of size 1 // 1 square (s) of size 2 // ********************************** int main(){ int kase = 1; while(scanf("%d", &n)!= EOF) { scanf("%d", &m); if(kase != 1)printf("**********************************\n"); memset(v,0,sizeof(v)); memset(h,0,sizeof(h)); memset(a,0,sizeof(a)); for( int i = 0; i < m; i++) { char s[5];int x1,y1; scanf("%s%d%d", s, &x1, &y1); if (s[0] == 'H') h[x1][y1] = 1; if (s[0] == 'V') v[y1][x1] = 1; getchar(); } for(int i = 1; i < 10; i++) for(int j = 1; j < 10; j++) panduan(i,j); printf("Problem #%d\n\n", kase++); int flag = 0; for(int i = 1; i < 10; i++) if(a[i] > 0) {flag=1;printf("%d square (s) of size %d\n", a[i], i);} if(!flag)printf("No completed squares can be found.\n"); } return 0; } /* H 横边 V 竖边 需要交换一下 x y 如题意 思路 用两个二维数组 存边 遍历点 每个点都判断下 1~9的方形 写个函数就是 try a try,ac is okay Sample Input 4 16 H 1 1 H 1 3 H 2 1 H 2 2 H 2 3 H 3 2 H 4 2 H 4 3 V 1 1 V 2 1 V 2 2 V 2 3 V 3 2 V 4 1 V 4 2 V 4 3 2 3 H 1 1 H 2 1 V 2 1 Sample Output Problem #1 2 square (s) of size 1 1 square (s) of size 2 ********************************** Problem #2 No completed squares can be found. 9 32 H 1 1 H 1 2 H 1 3 H 1 4 H 1 5 H 1 6 H 1 7 H 1 8 H 9 1 H 9 2 H 9 3 H 9 4 H 9 5 H 9 6 H 9 7 H 9 8 V 1 1 V 1 2 V 1 3 V 1 4 V 1 5 V 1 6 V 1 7 V 1 8 V 9 1 V 9 2 V 9 3 V 9 4 V 9 5 V 9 6 V 9 7 V 9 8 Problem #1 1 square (s) of size 8 */
习题 4-3 黑白棋 uva220
模拟黑白棋， 注意 若当前操控者走不了，就切换另一个操控者
解：类似之前的象棋，两个字符串数组 一个存储棋盘 一个存可能的走法 ，但是WA WA WA
// 黑白棋 8*8 的棋盘 横竖斜 #include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; char oth[9][9]; bool may[9][9]; char cur,wro; int x,y,cnt,flag; void init(char curplayer,char wa){ for(int i = 1; i < 9; i++)for(int j = 1; j < 9; j++)may[i][j] = false; cnt = 0; for(int i = 1; i < 9; i++) for(int j = 1; j < 9; j++) if(oth[i][j] == curplayer) { if(oth[i][j+1] == wa) for(int k = j+2; k < 9; k++) { if(oth[i][k] == '-') { may[i][k] = true;cnt++;break; } if(oth[i][k] == curplayer) break; } if(oth[i][j-1] == wa) for(int k = j-2; k > 0; k--) { if(oth[i][k] == '-') { may[i][k] = true;cnt++;break; } if(oth[i][k] == curplayer) break; } if(oth[i+1][j] == wa) for(int k = i+2; k < 9; k++) { if(oth[k][j] == '-') { may[k][j] = true;cnt++;break; } if(oth[k][j] == curplayer) break; } if(oth[i-1][j] == wa) for(int k = i-2; k > 0; k--) { if(oth[k][j] == '-') { may[k][j] = true;cnt++;break; } if(oth[k][j] == curplayer) break; } if(oth[i+1][j+1] == wa) for(int k = 2; i+k<9 && j+k<9; k++) { if(oth[i+k][j+k] == '-') { may[i+k][k+j] = true;cnt++;break; } if(oth[i+k][j+k] == curplayer) break; } if(oth[i-1][j-1] == wa) for(int k = 2; i-k>0 && j-k>0; k++) { if(oth[i-k][j-k] == '-') { may[i-k][j-k] = true;cnt++;break; } if(oth[i-k][j-k] == curplayer) break; } if(oth[i-1][j+1] == wa) for(int k = 2;i-k>0 && j+k<9; k++) { if(oth[i-k][j+k] == '-') { may[i-k][j+k] = true;cnt++;break; } if(oth[i-k][j+k] == curplayer) break; } if(oth[i+1][j-1] == wa) for(int k = 2;i+k<9 && j-k>0; k++) { if(oth[i+k][j-k] == '-') { may[i+k][j-k] = true;cnt++;break; } if(oth[i+k][j-k] == curplayer) break; } } } void bian(char curplayer,char wa){ int i = x,j = y; if(oth[i][j+1] == wa) for(int k = j+2; k < 9; k++) { if(oth[i][k] == '-')break; if(oth[i][k] == curplayer) { for(int l = k; l >= j; l--) oth[i][l] = curplayer; break; } } if(oth[i][j-1] == wa) for(int k = 2; j-k > 0; k++) { if(oth[i][j-k] == '-')break; if(oth[i][j-k] == curplayer) { for(int l = k; l >= 0; l--) oth[i][j-l] = curplayer; break; } } if(oth[i+1][j] == wa) for(int k = i+2; k < 9; k++) { if(oth[k][j] == '-')break; if(oth[k][j] == curplayer) { for(int l = k; l >= i; l--) oth[l][j] = curplayer; break; } } if(oth[i-1][j] == wa) for(int k = 2; i-k > 0; k++) { if(oth[i-k][j] == '-')break; if(oth[i-k][j] == curplayer) { for(int l = k; l >= 0; l--) oth[i-l][j] = curplayer; break; } } if(oth[i+1][j+1] == wa) for(int k = 2; i+k<9 && j+k<9; k++) { if(oth[i+k][j+k] == '-')break; if(oth[i+k][j+k] == curplayer) { for(int l = k; l >= 0; l--) oth[i+l][j+l] = curplayer; break; } } if(oth[i-1][j-1] == wa) for(int k = 2; i-k>0 && j-k>0; k++) { if(oth[i-k][j-k] == '-')break; if(oth[i-k][j-k] == curplayer) { for(int l = k; l >= 0; l--) oth[i-l][j-l] = curplayer; break; } } if(oth[i-1][j+1] == wa) for(int k = 2; i-k>0 && j+k<9; k++) { if(oth[i-k][j+k] == '-')break; if(oth[i-k][j+k] == curplayer) { for(int l = k; l >= 0; l--) oth[i-l][j+l] = curplayer; break; } } if(oth[i+1][j-1] == wa) for(int k = 2; j-k>0 && i+k<9; k++) { if(oth[i+k][j-k] == '-')break; if(oth[i+k][j-k] == curplayer) { for(int l = k; l >= 0; l--) oth[i+l][j-l] = curplayer; break; } } } int main() { int t;int t1 = t; cin>>t; while(t--){ if(t1 != t)getchar(); for(int i = 1;i <= 8; i++) { string s; cin>>s; for(int j = 0; j < 8; j++) oth[i][j+1] = s[j]; } getchar(); cur = getchar();wro = (cur == 'B'?'W':'B'); char cur1 = cur,wro1 = wro; init(cur,wro); while(1){ char ch = getchar(); if(ch == 'Q'){ for(int i = 1; i < 9; i++){for(int j = 1; j < 9; j++) putchar(oth[i][j]);putchar('\n');} if(t)printf("\n"); break; } if(ch == 'L'){ int cnt1 = 0; //printf("flag=%d cur=%c\n", flag, flag?cur:wro); //init(cur,wro); int f = 0; for(int i = 1; i < 9; i++)for(int j = 1; j < 9; j++) if(may[i][j]){ if(!f)printf("(%d,%d)", i, j);else printf(" (%d,%d)", i, j); f = 1;cnt1++; } if(f) printf("\n"); } if(ch == 'M'){ flag = 1; scanf("%1d%1d", &x, &y);if(!may[x][y]){ puts("No legal move."); flag = 0; bian(wro,cur); } else bian(cur,wro); //for(int i = 1; i < 9; i++){for(int j = 1; j < 9; j++)putchar(oth[i][j]);putchar('\n');} int b = 0,w = 0;for(int i = 1; i < 9; i++)for(int j = 1; j < 9; j++) {if(oth[i][j] == 'B')b++;if(oth[i][j] == 'W')w++;} printf("Black - %2d White - %2d\n", b, w); if(flag){swap(wro,cur);} init(cur,wro); //putchar(curplayer); // printf(" sdsad\n"); } } } return 0; } /* Sample Input 2 -------- -------- -------- ---WB--- ---BW--- -------- -------- -------- W L M35 L Q WWWWB--- WWWB---- WWB----- WB------ -------- -------- -------- -------- B L M25 L Q Sample Output (3,5) (4,6) (5,3) (6,4) Black - 1 White - 4 (3,4) (3,6) (5,6) -------- -------- ----W--- ---WW--- ---BW--- -------- -------- -------- No legal move. Black - 3 White - 12 (3,5) WWWWB--- WWWWW--- WWB----- WB------ -------- -------- -------- -------- */
习题 4-4 骰子涂色 uva253
输入一个长度12的字符串 分别表示两个正方体的上前左右后下 六个面
问是否为同一个正方体 （可通过旋转）
解：三对两两前后的面 分别与 另一正方形相同即可
//骰子涂色 touzituse #include<iostream> #include<cstdio> #include<string> using namespace std; int main(){ string s; while(cin >> s){ string s11,s12,s13,s21,s22,s23; s11 = s[0]+s[5]; s12 = s[1]+s[4]; s13 = s[2]+s[3]; s21 = s[6]+s[11]; s22 = s[7]+s[10]; s23 = s[8]+s[9]; //cout<< s << endl; //cout<< s11 << s12 << s13 << s21 << s22 << s23 << endl;//乱码？？？ 为啥输出不了这些个字符串 /*if((s11==s21 && s12==s22 && s13==s23)||(s11==s22 && s12==s23 && s13==s21)||(s11==s23 && s12==s21 && s13==s22) || (s11==s21 && s12==s23 && s13==s22)||(s11==s22 && s12==s21 && s13==s23)||(s11==s23 && s12==s22 && s13==s21))*/ //直接穷举可能的情况 六种而已 ，有解法反着又来一次 但好像没必要 int flag = 0; if ((s11==s21||s11==s22||s11==s23) && (s12==s21||s12==s22||s12==s23) && (s13==s21||s13==s22||s13==s23)) if ((s21==s11||s21==s12||s21==s13) && (s22==s11||s22==s12||s22==s13) && (s23==s11||s23==s12||s23==s13)) flag = 1; puts(flag?"TRUE":"FALSE"); } return 0; }
习题4-5 IP网络 uva1590
输入n组IP地址，问最小网络地址与子网掩码
解：找不同，可以直接用个二维数组 存每个IP地址的32位二进制 ，或者从左到右比较数（我的做法）
#include<algorithm>//IP 网络 #include<cstring> #include<cstdio> using namespace std; const int ma = 1000+10; int x,y,n; int a[ma],b[ma],c[ma],d[ma],maxx,minn; void fin(){ x = 0,y = 0;int s = 1<<7;//printf("maxx= %d minn= %d\n", maxx, minn); for(int i = 7; i >= 0; i--) { int t = (maxx/s)&1; if(((minn/s)&1) == t) x += t*s, y += s; else break; s /= 2; } } int main(){ while(scanf("%d",&n) != EOF) { memset(a,0,sizeof(a));memset(b,0,sizeof(b)); memset(c,0,sizeof(c));memset(d,0,sizeof(d)); for(int i = 1; i <= n; i++)scanf("%d.%d.%d.%d", &a[i], &b[i], &c[i], &d[i]); if(n == 1){printf("%d.%d.%d.%d\n255.255.255.255\n", a[1], b[1], c[1], d[1]);} else { sort(a+1,a+n+1);sort(b+1,b+n+1);sort(c+1,c+n+1);sort(d+1,d+n+1); if(a[1] == a[n]) { if(b[1] == b[n]) { if(c[1] == c[n]) { if(d[1] == d[n]) { printf("%d.%d.%d.%d\n",a[1], b[1], c[1], d[1]); printf("255.255.255.255\n"); } else { minn = d[1];maxx = d[n];fin(); printf("%d.%d.%d.%d\n", a[1], b[1], c[1], x); printf("255.255.255.%d\n", y); } } else { minn = c[1];maxx = c[n];fin(); printf("%d.%d.%d.0\n", a[1], b[1], x); printf("255.255.%d.0\n", y); } } else { minn = b[1];maxx = b[n];fin(); printf("%d.%d.0.0\n", a[1], x); printf("255.%d.0.0\n", y);//stupid fault } } else { minn = a[1];maxx = a[n];fin(); printf("%d.0.0.0\n", x); printf("%d.0.0.0\n", y); } } } return 0; }
习题4-6 莫尔斯密码 uva 508
单个字符编码，给出部分单词 字符串匹配 若多个精确匹配 输出第一个匹配的+！ 若无法精确匹配 输出最接近的+？
运用好的容器，事半功倍 自己想来想去都太复杂了 就找了别人的写法
#pragma GCC diagnostic error "-std=c++11" #include<iostream> #include<string>//No morse word will have more than eighty (80) elements. #include<cstdio>//Thered will be at most 100 context words #include<map> const int INF = 1e5; using namespace std; map<char, string> ascode; map<string, string> encod; int dist(string a, string b) { if(a.size() > b.size()) swap(a,b); if(a == b.substr(0, a.size())) return b.size()-a.size(); return INF; } string encode(string& s) { string ans; for(int i = 0; i < s.size(); i++) ans += ascode[s[i]]; return ans; } string decode(string s) { auto it = encod.begin(); string ans = it->first; //cout<<" || " << ans << endl; int dis = dist(it->second, s); while(++it != encod.end()) { int d = dist(it->second, s); if(d < dis) dis = d, ans = it->first; else if(d == dis && d == 0 && ans[ans.size()-1] != '!') ans += '!'; } if(dis != 0) ans += '?'; return ans; } int main() { string s1,s2; while(cin >> s1 && s1[0] != '*') { cin >> s2; ascode[s1[0]] = s2; //cout<< "ascode."<<s1[0] << "="<<ascode[s1[0]] <<endl; } while(cin >> s1 && s1[0] != '*') { encod[s1] = encode(s1); //cout<<s1<<"||"<< encode(s1)<<endl; } while(cin >> s1 && s1[0] != '*') { //cout<<endl<<s1<<"__"<<endl; cout<<decode(s1) << endl; } return 0; } /* Sample Input A .- B -... C -.-. D -.. E . F ..-. G --. H .... I .. J .--- K -.- L .-.. M -- N -. O --- P .--. Q --.- R .-. S ... T - U ..- V ...- W .-- X -..- Y -.-- Z --.. 0 ------ 1 .----- 2 ..--- 3 ...-- 4 ....- 5 ..... 6 -.... 7 --... 8 ---.. 9 ----. * AN EARTHQUAKE EAT GOD HATH IM READY TO WHAT WROTH * .--.....-- .....--.... --.----.. .--.-.----.. .--.....-- .--. ..-.-.-....--.-..-.--.-. ..-- .-...--..-.-- ---- ..-- * Sample Output WHAT HATH GOD WROTH? WHAT AN EARTHQUAKE EAT! READY TO EAT! */
xiti 4-7 RAID技术 uva 509
没看懂题。。。。
#include<bits/stdc++.h> //给出数据块，判断合法性，合法则恢复并输出完整的数据。非法则报告磁盘非法。 //合法： 1.任意一列如果没有x则应该满足奇校检或偶校检条件(即1的个数为奇数或者偶数)。 //2.某一列如果有x，只允许有一个。 using namespace std; char mp[8][6420]; int d, s, b;int op; char c; int input() { cin >> d; memset(mp, 0, sizeof(mp)); if(d == 0) return 0; cin >> s >> b; cin >> c; op = (c=='E')?0:1; for(int i = 0; i < d; i++) scanf("%s", mp[i]); return 1; } int isLegal() { for(int i = 0; i < s*b; i++) { int tag = 0, numx = d, cntx = 0; for(int j = 0; j < d; j++) { if(mp[j][i] == 'x') { if(cntx == 0) { cntx++; numx = j; } else return 0; } else { if(j == 0) tag = mp[j][i] - '0'; else tag ^= (mp[j][i] - '0'); } } if(cntx == 0 && tag != op) return 0; else mp[numx][i] = (tag ^ op) + '0'; } return 1; } void recover() { string str = ""; for(int i = 0; i < b; i++) { for(int j = 0; j < d; j++) { //printf("i=%d j=%d\n", i, j); if(j == i % d) continue; for(int k = i*s; k < (i+1)*s; k++) { str += mp[j][k]; //printf("i=%d j=%d k=%d\n", i, j, k); }//cout<< str <<endl; //printf(" i=%d j=%d\n", i, j); } } int tag = 0; //cout<< str.length() << endl; while(str.length() % 4) str += '0'; //cout<< str <<endl; for(int i = 0; i < str.length(); i += 4) { int t = 0; for(int j = 0; j < 4; j++) { t *= 2; t += (str[i+j] - '0'); //cout<< i<< "_" << j << " "<<str[i+j] <<"__"<<endl; } printf("%X", t); } cout << endl; } int main() { //freopen("D:\\input.txt", "r", stdin); //freopen("D:\\output.txt", "w", stdout); //ios::sync_with_stdio(false); int kase = 0; while(input()) { if(!isLegal()) printf("Disk set %d is invalid.\n",++kase); else{ printf("Disk set %d is valid, contents are: ",++kase); recover(); } /*for(int i = 0; i < s*b; i++) { for(int j = 0; j < d; j++) printf("%c", mp[j][i]); printf("\n"); }*/ } return 0; }
习题4-8 特别困的学生 uva 12108
暴力模拟 、
#include<cstdio> #define N 298 int main(){ int n, a[15], b[15], c[15];int cnt, kase = 1; while(scanf("%d", &n) , n){ int sleep = 0,count = 0; for(int i = 0; i < n; i++) scanf("%d%d%d", &a[i], &b[i], &c[i]); for(cnt = 1; cnt < N; cnt++){ count = 0; for(int j = 0; j < n; j++){ if(c[j] <= a[j]) count++; } if(count == n)break; for(int j = 0; j < n; j++){ if(c[j]==(a[j]+b[j]) || (c[j]==a[j] && n-count <= count)) c[j] = 0; c[j]++; //printf("count=%d c[%d]=%d\n", count, j, c[j]); } } if(cnt == N)cnt = -1; printf("Case %d: %d\n", kase++, cnt); } return 0; }
习题 4-9 数据挖掘 uva1591
题没看懂。。。
习题 4-10 洪水 uva 815
//题意 可理解为 n*m个山峰， 给定水的体积， 问水能淹过几个山峰，并且最后水位是？
//比如说 给定 3，4，5，6， 7， 8， 9， 11， 12 山峰， 水体积900（底面积100，so 可用高度为9 ）
//水只能淹过前面四个山峰， 水平面变成 6+（9-（6*4-3-4-5-6））/4 = 6.75
//有水区域占比则为 4/9 = 44.44%
//将输入的山峰高度 排个序， 有个小技巧 在最高的山峰后面加一个特别大的山峰
//程序实现如下 #include<algorithm> #include<iostream> #include<cstring> #include<cstdio> using namespace std; int main(){ int n,m,a[920],kase = 1; double sum,he,x,he1; while(scanf("%d%d", &n, &m), n) { n *= m; memset(a, 0, sizeof(a)); for(int i = 0; i < n; i++) scanf("%d", a+i); sort(a, a+n); scanf("%lf", &sum); sum /= 100.0; he = 0; int c; a[n] = 0x3f3f3f3f;//??? for(c = 1; c < n; c++)// == n { he += (a[c]-a[c-1])*c; if(he >= sum) break; } for(int i = 0; i < c; i++) sum += a[i]; printf("Region %d\n", kase++); printf("Water level is %.2lf meters.\n%.2lf percent of the region is under water.\n\n", (sum)*1.0/c, (c)*100.0/n); } return 0; } /* Sample Input 3 3 25 37 45 51 12 34 94 83 27 10000 0 0 Sample Output Region 1 Water level is 46.67 meters. 66.67 percent of the region is under water. */
#include <algorithm> #include <cstdio> using namespace std; const int maxn = 35; int a[maxn*maxn], n, m; int main() { for (int kase = 1; scanf("%d%d", &n, &m), n&&m; kase++) { n *= m; for(int i = 0; i < n; i++) scanf("%d", a+i); sort(a, a+n); a[n] = 0x3f3f3f3f; int k; double sum; scanf("%lf", &sum); sum /= 100.0; for (k = 1; k <= n; k++) { sum += a[k-1]; if (sum/k <= a[k]) break; } printf("Region %d\n", kase); printf("Water level is %.2lf meters.\n", sum/k); printf("%.2lf percent of the region is under water.\n\n", k*100.0/n); } return 0; }
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