随笔分类 -  UvaOJ

摘要：SPFA，2RE，无向图的边数为题目描述的两倍。# include <cstdio># include <cstring># include <queue>using namespace std;# define N 20000 + 5# define M (50000 + 5) * 2# define INF 550000000int n, m, src, des;int first[N], d[N];bool inq[N];int u[M], v[M], w[M], next[M];void read_graph(void){ scanf("% 阅读全文

摘要：时限是 3s， 当然不能用 dijksra；N 最大是 20005，数组 d[N][N] 可以开；------------------------------------------------------------------------# include <stdio.h># include <string.h># define N 20005# define M 50005# define INF N*Mint n, m, s, t;int w[N][N];int min(int x, int y){ return (x<y ? x:y);}void in 阅读全文

摘要：这道题重点在于理解题意，不是动态规划？（If an organism were in two stages of growth at the same time the first option from the list above should be given as an answer.）：初始阶段微生物为A，有两种生长方式，一种是右边扩展“AB”，一种是最前面扩展“A”而最后面扩展“B”3；所以合法的微生物的长度都是奇数，然后再判断是否是以上三种阶段即可；参考了网上的分析。 1 # include <stdio.h> 2 # include <string.h> 阅读全文

摘要：复习了the C programming language 中的单词记数/*494 - Kindergarten Counting Game*/# include <stdio.h># include <ctype.h># define IN 1# define OUT 0int main(){ int ans,state; char ch; ans = 0; state = OUT; while ((ch=getchar()) != EOF) { if (!isalpha(ch) && ch!='\n') state =... 阅读全文

摘要：看出来了，ascll -7WA：/*458 - The Decoder*/# include <stdio.h>int main(){ char ch; while ((ch=getchar()) != EOF) if (ch != '\n') putchar((ch+121)%128); else putchar('\n'); return 0;}AC：/*458 - The Decoder*/# include <stdio.h>int main(){ char ch; while ((ch=getchar()) != EOF) .. 阅读全文

摘要：理解错误：Do not output any blank lines.注意： No integer in the input is greater than 100000 or less than 0./*10300 - Ecological Premium*/# include <stdio.h>int main(){ int n, f; long long s, a, e, ans; scanf("%d", &n); while (n > 0) { ans = 0; scanf("%d", &f); while (. 阅读全文

摘要：初速度与加速度恒定，已知经 t 时间后速度为 v， 求经 2t 后位移为多少？/*10071 - Back to High School Physics*/# include <stdio.h>int main(){ int v0, t; while (scanf("%d%d", &v0, &t)!=EOF) printf("%d\n",2*v0*t); return 0;} 阅读全文

摘要：3WA。。。/* UVa10055 Hashmat */# include <stdio.h>int main(){ unsigned long a, b; while (scanf("%ld%ld", &a, &b) != EOF) printf("%ld\n", (a<b ? b-a:a-b)); return 0;} 阅读全文