1048. Find Coins (25) Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special require…

1048. Find Coins (25) 时间限制 50 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept a…

1048 Find Coins (25 分)   Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requ…

时间限制 50 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as p…

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: f…

题目 Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment…

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: f…

先对序列排序,然后枚举较小值,二分较大值. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<vector> using namespace std; +; i…

给n,m以及n个硬币 问,是否存在两个硬币面值v1+v2=m 因为面值不会超过500,所以实际上最多500个不同的硬币而已 #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> using namespace std; +; int coin[maxn]; int n,m; int main() { scanf("%d %d",&…

题意: 输入两个正整数N和M(N<=10000,M<=1000),然后输入N个正整数(<=500),输出两个数字和恰好等于M的两个数(小的数字尽可能小且输出在前),如果没有输出"No Solution". AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; int main(){ ios::sync_with_stdi…