平时十测
题解：
第一题：大模拟，横躺竖躺站立三种状态，对每种状态分别模拟就好了；
我记录每面旋转后对应原来的面，很明显麻烦了，我大模拟写了2个小时，别人40分钟A，DS能力太差了；
#include<bits/stdc++.h> using namespace std; const int M = 2005; char opt[105]; int vis[M][M], mx; int zz[8]; int zl[5][2] = {{0,0},{1,0},{-1,0},{0,1},{0,-1}}, a[2]; inline int get(char c){ switch(c){ case 'E':return 1; case 'W':return 2; case 'N':return 3; case 'S':return 4; } } int ff[8], p[8]; int sta[7][8] = { {0, 1, 2, 6, 5, 3, 4}, {0, 5, 2, 6, 4, 3, 1},//E {0, 6, 2, 5, 4, 1, 3},//W {0, 2, 3, 4, 1, 5, 6},//N {0, 4, 1, 2, 3, 5, 6}//S }; void init(){ memset(vis, 0, sizeof(vis)); mx = 0; for(int i=1;i<=6;i++)zz[i]=sta[0][i]; } inline void work(int t){ for(int i=1;i<=6;i++)ff[i]=zz[i]; for(int i=1;i<=6;i++)zz[i]=ff[sta[t][i]]; } int main(){ freopen("block.in","r",stdin); freopen("block.out","w",stdout); int T, h; scanf("%d", &T); p[1] = p[6]= 1; while(T--){ init(); scanf("%d", &h); for(int i=2;i<=5;i++)p[i]=h; scanf("%s", &opt); int st = 1000, ed = 1000, ss = 1000, dd = 1000; vis[st][ed]++; int len = strlen(opt); for(int i = 0; i < len; i++){ int t = get(opt[i]); int dx = zl[t][0], dy = zl[t][1], nx, ny, now; if(dx){ if(p[zz[6]]==1&&p[zz[1]]==h)now=1; else if(p[zz[6]]==p[zz[1]]&&p[zz[1]]==h)now=1; else now=h; if(dx > 0){ nx = ss + dx * now; for(int k=ed;k<=dd;k++) for(int j=ss+1;j<=nx;j++)if( (++vis[j][k])>mx ) mx = vis[j][k]; } if(dx < 0) { nx = st + dx * now; for(int k=ed;k<=dd;k++) for(int j=st-1;j>=nx;j--)if( (++vis[j][k])>mx ) mx = vis[j][k]; } } else{ if(p[zz[6]]==1&&p[zz[1]]==h)now=1; else if(p[zz[6]]==p[zz[1]]&&p[zz[1]]==h)now=1; else now=h; if(dy > 0){ ny = dd + dy * now; for(int k=st;k<=ss;k++) for(int j=dd+1;j<=ny;j++)if( (++vis[k][j])>mx ) mx = vis[k][j]; } if(dy < 0) { ny = ed + dy *now; for(int k=st;k<=ss;k++) for(int j=ed-1;j>=ny;j--)if( (++vis[k][j])>mx ) mx = vis[k][j]; } } if(dx){ if(dx > 0) { st=ss+1;ss=nx; //if(ss==st) } else { ss=st-1;st=nx; } } else{ if(dy > 0){ ed=dd+1;dd=ny; } else { dd=ed-1;ed=ny; } } //printf("%d:%c %d %d %d %d %d %d %d %d %d\n",i, opt[i],st,ss,ed,dd,dx,dy,now,nx,ny); work(t); //for(int i=1;i<=6;i++)printf("%d ",zz[i]);puts(""); } int sx=max(ss-st+1, dd-ed+1); if(ss==st) for(int i=1;i<=sx;i++)printf("%d ", ss-1000); else for(int i=st;i<=ss;i++)printf("%d ", i-1000); puts(""); if(ed==dd)for(int i=1;i<=sx;i++)printf("%d ",dd-1000); else for(int i=ed;i<=dd;i++)printf("%d ", i-1000); puts(""); printf("%d\n",mx); } }
第二题：前面O（N）50%的分，后面就不会了；
#include<bits/stdc++.h> using namespace std; #define ll long long #define RG register ll bin[20]; const int M = 1e7+2; int bit[M][4]; inline void cut(ll &num){ while(num%10 == 0)num/=10; num%=bin[9]; } void init(){ ll now = 1; bin[0] = 1; for(int i = 1; i <= 17; i++) bin[i]=bin[i-1]*10; for(RG ll i = 1; i <= 1e7; i++){ now*=i; cut(now); for(int j = 1; j <= 3; j++){ bit[i][j] = now%bin[j]; } // printf("%lld %d\n", i, bit[i][3]); } } void print(int k, int num){ if(k == 3)printf("%03d\n",num); if(k == 2)printf("%02d\n",num); if(k == 1)printf("%d\n",num); } int main(){ freopen("num.in","r",stdin); freopen("num.out","w",stdout); int T; scanf("%d", &T); //int tt=clock(); init(); while(T--){ int n, k; scanf("%d%d", &n, &k); print(k, bit[n][k]); } //int cc=clock(); ////fprintf(stderr,"FUCK %d",cc-tt); }
第三题：对于每个节点答案组成必须是1234……k，才合法，我们可以每次在节点后面添加id，最后判断组成串是否合法即可，这就相当于一个哈希，变成了经典的线段树操作，区间乘一个数加一个数
#include<bits/stdc++.h> using namespace std; const int M = 200005; #define ll unsigned long long const ll bas = 2333; int n, k, cnt; ll stan; struct Node{ Node *ls, *rs; ll A, B; void down(){ if(A == 1 && !B)return ; ls->B = ls->B*A + B; ls->A *= A; rs->B = rs->B*A + B; rs->A *= A; A = 1, B = 0; } }pool[M<<2], *root,*tail=pool; Node *build(int lf=1,int rg=n){ Node *nd =++tail; nd->A = 1, nd->B = 0; if(lf==rg); else { int mid=(lf+rg)>>1; nd->ls=build(lf,mid); nd->rs=build(mid+1,rg); } return nd; } #define Ls lf,mid,nd->ls #define Rs mid+1,rg,nd->rs void modify(int L, int R, int dv, int lf=1,int rg=n,Node *nd = root){ if(L <=lf && rg <= R){ nd->A *= bas; nd->B = nd->B * bas + dv; } else { int mid=(lf+rg)>>1; nd->down(); if(L<=mid)modify(L, R, dv, Ls); if(R>mid)modify(L, R, dv, Rs); } } void getdown(int lf=1,int rg=n,Node *nd=root){ if(lf==rg){ if(nd->B == stan) cnt++; } else { int mid=(lf+rg)>>1; nd->down(); getdown(Ls); getdown(Rs); } } int main(){ freopen("deco.in","r",stdin); freopen("deco.out","w",stdout); int T; scanf("%d%d%d", &n,&k,&T); root=build(); int l,r,id; for(int i=1;i<=T;i++){ scanf("%d%d%d",&l,&r,&id); modify(l,r,id); } stan = 0; for(int i = 1; i <= k; i++) stan = stan * bas + i; getdown(); printf("%d\n",cnt); }
