比较坑的一道题...直接做肯定WA http://acm.hdu.edu.cn/showproblem.php?pid=1013 #include<iostream> #include<cstring> using namespace std; int main(){ ]; while(cin>>a){ int len = strlen(a); ; ;i<len;i++){ n += a[i]-'; } ) break; ){ ; while(n ){ sum +…

Problem Description The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits a…

HDOJ 题目分类 //分类不是绝对的 //"*" 表示好题,需要多次回味 //"?"表示结论是正确的,但还停留在模块阶 段,需要理解,证明. //简单题看到就可以敲的 1000:    入门用: 1001:    用高斯求和公式要防溢出 1004:1012: 1013:    对9取余好了 1017:1021: 1027:    用STL中的next_permutation() 1029:1032:1037:1039:1040:1056:1064:1065: 10…

HDOJ 题目分类 /* * 一:简单题 */ 1000:    入门用:1001:    用高斯求和公式要防溢出1004:1012:1013:    对9取余好了1017:1021:1027:    用STL中的next_permutation()1029:1032:1037:1039:1040:1056:1064:1065:1076:    闰年 1084:1085:1089,1090,1091,1092,1093,1094, 1095, 1096:全是A+B1108:1157:1196:1…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…

1001. A+B Format (20) 注意负数,没别的了. 用scanf来补 前导0 和 前导的空格 很方便. #include <iostream> #include <cstdio> using namespace std; ]; int main() { int A,B; cin>>A>>B; A+=B; ) { A=-A; cout<<"-"; } ; while(A) { a[n++]=A%; A/=; } ;…

Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…

卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…