ZOJ 1711 H-Sum It Up】的更多相关文章

Sum It Up Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 4   Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Given a specified total t and…
题目链接:http://poj.org/problem?id=1564 题目大意:给定一个整数t,和n个元素组成的集合.求能否用该集合中的元素和表示该整数,如果可以输出所有可行解.1<=n<=12 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…
POJhttp://poj.org/problem?id=1564 ZOJhttp://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=711 HDUhttp://acm.hdu.edu.cn/showproblem.php?pid=1258 今天不好玩.学校网断了,wifi也用不了,爪机当热点还上不去!!!A一题容易吗!然后鼠标更不好使了,果断去买了一个. 瞬间又成穷人.T T 换了个鼠标心情大好~ -------------------…
Description John von Neumann, b. Dec. 28, 1903, d. Feb. 8, 1957, was a Hungarian-American mathematician who made important contributions to the foundations of mathematics, logic, quantum physics,meteorology, science, computers, and game theory. He wa…
https://vjudge.net/contest/67836#problem/H 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…
1.http://blog.csdn.net/dyx404514/article/details/39122743 思路:题目意思很清楚了,这里只说思路. 设区间[L,R],区间长度为len=(R-L+1),设F[i]表示G(i,i)+G(i,i+1)+......G(i,R).那么对于区间[L,R],询问的答案即为Ans[L,R]=F[L]+F[L+1]+......+F[R].容易得到G[i,j]=Si-G(i+1,j),所以F[i]=(R-i+1)*Si-F[i+1].即F[i]+F[i+…
题目:click here #include <bits/stdc++.h> using namespace std; typedef unsigned long long ll; const int INF = 0x3f3f3f3f; ; int n; int a[M][M]; // a[i][j] 表示从[i][0]到[i][j]的和 int main() { while( ~scanf("%d", &n ) ) { memset( a, , sizeof(a)…
Problem H {sum+=i++} to Reach N Input: standard input Output:  standard output Memory Limit: 32 MB All the positive numbers can be expressed as a sum of one, two or more consecutive positive integers. For example 9 can be expressed in three such ways…
双塔DP. dp[i][j]表示前i个物品,分成两堆(可以不全用),价值之差为j的时候,较小一堆的价值为dp[i][j]. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; ][ + ]; ]; int n, sum; void read() { ; i <= n; i++) scanf("%d",…
Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.Example 1:Input:nums = [1,1,1], k = 2Output: 2Note:The length of the array is in range [1, 20,000].The range of numbers in th…