迎接2012新赛季——HDOJ系列热身赛（5） 部分题目
今天下午航电的比赛，我没注意，下午两点多才来实验室。只是做了几道水题。
4229
就是给定那个a,b,c,d四个非负整数数，分别用|a-b| |b-c| |c-d| |d-a| 来代替，计算直到 a= = b == c == d为止求出一共需要多少步，这题幸亏给出了 It turns out that in all cases, the sequence converges very quickly. In fact, it can be shown that if all four integers are less than 2^n, then it will take no more than 3*n steps to converge! 这句话所以才刚直接做，如果没有这句的话，我想很少有人敢这样写了吧。按题意模拟记录步数即可。
4235
给定三个数v = a*b 其中形成a,b的数是由V中的元素组成的。求大于等于给定x的满足上面条件的数；
求出所有的数然后打表处理；
View Code
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <cmath> #define maxn 1000001 using namespace std; //int f[maxn],L,t[10]; int path[10]; bool hash[10],vt[10]; bool h[maxn + 99999]; int cmp(int a,int b) { return a < b; } int f[738] = { 126,153,688,1206,1255,1260,1395,1435,1503,1530,1827,2187,3159,3784,6880,10251,10255,10426,10521,10525,10575,11259,11844,11848,12006,12060,12384,12505,12546,12550,12595,12600,12762,12843,12955,12964,13243,13545,13950,14035,14350,15003,15030,15246,15300,15435,15624,15795,16272,17325,17428,17437,17482,18225,18265,18270,19026,19215,21375,21586,21753,21870,25105,25375,25474,25510,28476,29632,31509,31590,33655,33696,36855,37840,37845,39784,41665,42898,44676,45684,45760,45864,47538,48672,49855,51759,52168,53865,56295,56875,62968,63895,67149,67392,67950,68800,71199,78975,100255,100525,101299,102051,102505,102510,102541,102550,102595,102942,102955,103968,104026,104260,104368,105021,105025,105075,105210,105250,105264,105295,105723,105750,107163,107329,108126,108135,108216,108612,108864,109525,110758,112468,112509,112590,114268,115672,115699,116478,116496,116725,116928,117067,118408,118440,118480,118575,118926,119848,120006,120060,120384,120600,120762,120843,121086,121576,121815,122746,122764,123084,123354,123538,123840,123894,124483,124488,124542,124978,125005,125050,125095,125248,125433,125460,125500,125950,125995,126000,126027,126108,126846,127417,127620,128403,128430,128943,129438,129505,129514,129550,129564,129595,129640,129775,129955,131242,132430,132565,132615,132655,133245,134275,134725,135045,135450,135828,135837,136525,136854,136948,138784,139500,139824,140035,140350,141345,142978,143500,143739,143793,145273,145314,145345,145683,146137,146520,146952,149364,149782,150003,150030,150246,150300,150435,150624,150826,152271,152406,152460,152608,152685,152946,153000,153436,154350,155277,156024,156240,156289,156325,156915,157950,158193,162072,162526,162720,162976,163255,163795,163854,163944,164583,165208,168520,171598,172246,172386,172822,173250,173925,174028,174082,174208,174280,174298,174370,174793,174802,174820,174982,175329,176215,178294,178942,179325,179428,179482,180225,180297,180621,182065,182250,182650,182700,182974,184126,186624,187029,189702,189742,190260,190827,191205,192150,192375,192685,192717,193257,193945,194229,197428,197482,197725,201852,205785,207391,208624,210375,210681,210753,211896,212868,213075,213466,213750,213759,214506,215086,215424,215455,215860,216733,217503,217530,217638,217854,218488,218700,223524,226498,226872,226876,227448,229648,231579,231673,233896,236754,236758,236925,238968,241506,241564,243175,245182,245448,246150,246928,250105,250510,251005,251050,251095,251896,253750,254740,255010,255100,256315,256410,256414,258795,259510,260338,261378,261783,262984,263074,263155,263736,267034,268398,279328,281736,283198,283648,284598,284760,285376,286416,286974,287356,289674,291375,291753,293625,295105,295510,296320,297463,297832,304717,307183,312475,312565,312655,312975,314199,314743,315009,315090,315490,315594,315625,315900,316255,319059,319536,325615,326155,326452,328419,328864,329346,329656,336195,336550,336960,338296,341284,341653,342688,346288,346725,346968,347913,352966,355995,361989,362992,365638,368104,368550,368784,369189,371893,373864,375156,375615,376992,378400,378418,378450,381429,384912,384925,386415,390847,392566,393246,393417,394875,397840,399784,404932,404968,414895,415575,416065,416259,416650,416988,419287,428980,429664,435784,439582,442975,446760,446976,447916,449676,449955,450688,451768,456840,457168,457600,458640,462672,465088,465984,468535,475380,475893,476892,486720,488592,489159,489955,490176,491688,493857,495328,497682,498550,515907,516879,517509,517590,519745,520168,520816,521608,521680,526792,529672,530379,531297,535968,536539,538650,549765,559188,562950,564912,567648,568750,571648,573768,588676,611793,611878,612598,614965,617728,618759,623758,629680,632875,638950,649638,661288,665919,667876,671409,671490,671944,673920,678892,679500,687919,688000,692712,697248,702189,702918,710496,711099,711909,711990,715959,719199,729688,736695,738468,741928,769792,773896,778936,782896,785295,789250,789525,789750,791289,792585,794088,798682,809919,809937,809964,815958,829696,841995,859968,899019,936985,939658,960988,1000255,1000525,1002501,1002505,1002550,1002595,1002955,1004251,1005025,1005201,1005250,1005295,1008126,1009525,1012099,1012198,1012297,1012396,1012495,1012581,1012594,1012693,1012792,1012891,1012990,1014975,1016568,1016635,1017382,1019722,1020051,1020510,1020537,1021968,1021999,1024555,1024582,1024884,1025005,1025041,1025046,1025050,1025095,1025100,1025410,1025500,1025779,1025950,1025995,1029042,1029420,1029505,1029550,1029595,1029955,1031899,1032565,1032655,1032876,1039680,1040026,1040260,1040368,1041799,1042353,1042542,1042600,1043680,1043968,1045296,1046209,1050021,1050025,1050075,1050210,1050250,1050295,1050750,1051263,1051533,1051699,1052100,1052235,1052500,1052640,1052950,1052995,1053153,1053265,1053315,1057108,1057162,1057230,1057500,1058715,1058890,1061599,1062247,1062517,1063251,1064254,1065325,1065748,1066572,1067166,1068192,1068592,1071063,1071499,1071630,1071958,1072512,1072539,1073029,1073290,1077219,1079239,1079653,1080126,1080135,1080216,1080612,1080864,1081206,1081260,1081350,1081399,1082160,1083285,1083537,1083942,1086012,1086120,1086129,1088640,1088964,1089612,1089913,1090525,1091299,1092816,1095025,1095250,1095295,1098238,1098522,1099525 }; /*void init() { int tmp[10]; int i,j; memset(h,false,sizeof(h)); for (i = 1; i <= 1000000 + 99999; ++i) { for (j = 1; j < sqrt(i); ++j) { int a = j, b = i / j; int l1 = 0, l2 =0; int c = i; if (a*b == c) { while (a != 0) { tmp[l1++] = a%10; a /= 10; } while (b != 0) { tmp[l1++] = b%10; b /= 10; } while (c != 0) { t[l2++] = c%10; c /= 10; } if (l1 == l2) { int ct = 0,k1,k2; memset(hash,false,sizeof(hash)); for (k1 = 0; k1 < l1; ++k1) { for (k2 = 0; k2 < l2; ++k2) { if (tmp[k1] == t[k2] && !hash[k2]) { hash[k2] = true; ct++; break; } } } //printf(">>>>>>%d\n",ct); if (ct == l1) { //printf("%d %d %d\n",j,i/j,i); int sum = 0; for (int k = l1 - 1; k >= 0; --k) sum = sum*10 + t[k]; if (!h[sum]) { h[sum] = true; f[L++] = sum; } } } } } } sort(f,f + L,cmp); printf("%d\n",L); for (i = 0; i < L; ++i) printf("%d,",f[i]); }*/ int main() { //freopen("data.in","r",stdin); //L = 0; //init(); int n; while (scanf("%d",&n),n) { for (int i = 0; i < 738; ++i) { if (f[i] >= n) { printf("%d\n",f[i]); break; } } } return 0; }
4226 DP
读了很久才读懂了题意；题意是给定路径的有n段，m条路，求按指定路径走所用的最短距离；
dp[i][j] 表示地i段走第j条路的最短距离；
对于l,r他是不能换路所以有dp[i][j] = dp[i - 1][j] + w;
二对于s要求出他可以换路的范围在这个范围内求出最短路径。 （才开始自己没有用100求换路的上下界导致wa了还几次）
dp[i][j] = min(dp[i][j],dp[i - 1][j1] + w) （j1 >= s && j1 <= e）
注意k表示的是最内边与圆心的距离。
View Code
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <cmath> #define N 1007 #define M 12 using namespace std; const double pi = acos(-1.0); double dp[N][M]; const double inf = 999999999.0; int main() { //freopen("data.in","r",stdin); int n,m,i,j; double k; char op[2]; while (scanf("%d%d",&n,&m)) { if (!n && !m) break; //printf("%d %d\n",n,m); for (i = 0; i < N; ++i) { for (j = 0; j < M; ++j) { if (i == 0) dp[i][j] = 0; else dp[i][j] = inf; } } for (i = 1; i <= n; ++i) { scanf("%s%lf",op,&k); //printf("%c %.2lf\n",op[0],k); //对于L，R只能有上一路径往下走 if (op[0] == 'R') { for (j = 1; j <= m; ++j) { dp[i][j] = dp[i - 1][j] + (pi*((j - 1)*10 +5 + k))/2.0; } } else if (op[0] == 'L') { for (j = 1; j <= m; ++j) { dp[i][j] = dp[i - 1][j] + (pi*((m - j)*10 +5 + k))/2.0;//L时注意路径的选取 } } else { int s,e; for (j = 1; j <= m; ++j) { //就是由于这一段范围贡献了好几次WA int len = (k/100); if (j - len > 1) s = j - len; else s = 1; if (j + len > m) e = m; else e = j + len; for (int j1 = s; j1 <= e; ++j1) { double s = (j - j1)*10; double w = sqrt(k*k + s*s); dp[i][j] = min(dp[i][j],dp[i - 1][j1] + w); } } } /*for (j = 1; j <= m; ++j) printf("%.2lf ",dp[i][j]); printf("****************\n");*/ } double Min = dp[n][1]; for (i = 2; i <= m; ++i) { Min = min(Min,dp[n][i]); } printf("%.2lf\n",Min); } return 0; }
