Problem Description Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t=4, n=6, and the list is [4,3,2,2,1,1], then there are four different sums that equal 4: 4,3…

Sum it up 题意:给定一个数sum,和n个数,求sum可以由这n个数里面的那几个数的和表示. Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t=4, n=6, and the list is [4,3,2,2,1,1], then there are four d…

由这两题可知拓扑排序是通过“小于”关系加边建图的 hdu2647 /* 拓扑排序的原则是把“小于”看成有向边 此题反向建图即可 并且开num数组来记录每个点的应该得到的权值 */ #include<bits/stdc++.h> #include<queue> using namespace std; #define maxn 20000 struct Edge{ int to,nxt; }edge[maxn<<]; int head[maxn],tot,n,m; voi…

给你两个数t,n 接下来输入n个数字 让你输出所有数字相加等于n的组合 4 6 4 3 2 2 1 1 t n 4 3+1 2+2 2+1+1 Sample Input 4 6 4 3 2 2 1 1 5 3 2 1 1 400 12 50 50 50 50 50 50 25 25 25 25 25 25 0 0 Sample Output Sums of 4: 4 3+1 2+2 2+1+1 Sums of 5: NONE Sums of 400: 50+50+50+50+50+50+25+2…

Sum It Up Problem Description Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t=4, n=6, and the list is [4,3,2,2,1,1], then there are four different sums that eq…

这两道题都需要用到拓扑排序,所以先介绍一下什么叫做拓扑排序. 这里说一下我是怎么理解的,拓扑排序实在DAG中进行的,根据图中的有向边的方向决定大小关系,具体可以下面的题目中理解其含义 Educational Codeforces Round 25  E. Minimal Labels 题目链接:http://codeforces.com/contest/825/problem/E 题意:给你一个有向无环图(DAG),有n个顶点,m条边,顶点的序号分别是1-n,现在给你1-n的数对n个顶点进行赋值…

#include<bits/stdc++.h> using namespace std; typedef unsigned int ui; typedef long long ll; typedef unsigned long long ull; #define pf printf #define prime1 1e9+7 #define prime2 1e9+9 #define scand(x) scanf("%llf",&x) #define f(i,a,b)…

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I - 深搜 基础 Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit Status Description Given a specified total t and a list of n integers, find all distinct sums using numbers from the…