lightoj1125:Divisible Group Sums(类01背包)
来源:互联网 发布:ubuntu下好玩的游戏 编辑:程序博客网 时间:2024/05/22 04:21
Given a list of N numbers you will be allowed to choose any M of them. So you can choose in NCM ways. You will have to determine how many of these chosen groups have a sum, which is divisible by D.
Input
Input starts with an integer T (≤ 20), denoting the number of test cases.
The first line of each case contains two integers N (0 < N ≤ 200) and Q (0 < Q ≤ 10). Here N indicates how many numbers are there and Q is the total number of queries. Each of the next N lines contains one 32 bit signed integer. The queries will have to be answered based on these N numbers. Each of the next Q lines contains two integers D (0 < D ≤ 20) and M (0 < M ≤ 10).
Output
For each case, print the case number in a line. Then for each query, print the number of desired groups in a single line.
Sample Input
Output for Sample Input
2
10 2
1
2
3
4
5
6
7
8
9
10
5 1
5 2
5 1
2
3
4
5
6
6 2
Case 1:
2
9
Case 2:
1
题意:从N个数选M个使他们的和能被D整除,求方案数。
思路:dp[i][j]表示选i个数,他们的和取余D的值为j的方案数,dp[i][j] += dp[i-1][(D+ j-(a[k]%D)%D)]
# include <stdio.h># include <string.h>int main(){ int a[203], t, T, i, j, k, D, M, N, n, Q, num; long long dp[11][21]; scanf("%d",&T); for(t=1; t<=T; ++t) { scanf("%d%d",&N,&Q); for(i=1; i<=N; ++i) scanf("%d",&a[i]); printf("Case %d:\n",t); while(Q--) { scanf("%d%d",&D,&M); memset(dp, 0, sizeof(dp)); dp[0][0] = 1; for(i=1; i<=N; ++i) for(j=M; j>0; --j) { num = a[i]%D; for(k=0; k<D; ++k) dp[j][k] += dp[j-1][(k-num+D)%D]; } printf("%lld\n",dp[M][0]); } } return 0;}
- lightoj1125:Divisible Group Sums(类01背包)
- lightoj 1125 - Divisible Group Sums 01背包变形
- 10616 - Divisible Group Sums(dp背包)
- uva 10616 - Divisible Group Sums(计数)
- 1125 - Divisible Group Sums (DP)
- LightOJ 1125 Divisible Group Sums (dp)
- Divisible Group Sums LightOJ
- UVA 10616 Divisible Group Sums
- UVa:10616 Divisible Group Sums
- UVA10616 - Divisible Group Sums(dp)
- UVa 10616 - Divisible Group Sums
- lightoj 1125 - Divisible Group Sums
- Light oj 1125 - Divisible Group Sums(简单dp)
- Light OJ 1125 Divisible Group Sums (DP)
- Light OJ 1125 Divisible Group Sums 背包DP 2017/1/23
- UVa 10616 Divisible Group Sums (DFS&DP)
- LightOJ - 1125 Divisible Group Sums(DP)
- LightOJ1125
- Cygwin的安装和使用
- hdu 1856 More is better
- BZOJ 2194 快速傅立叶之二
- Javascript 中 Y 组合子的推导
- tessorflow练习 手写汉字识别
- lightoj1125:Divisible Group Sums(类01背包)
- javascript 基础 图片库和表单处理
- 执行应用程序出现: No such file or directory
- 计网学习笔记——概述
- 闲聊百度网盘api(秒传,离线,文件管理)
- 安利一款数学公式编辑器,AxMath
- 安全框架Shiro和Spring Security比较
- UVa120 Stacks of Flap jacks
- Mybatis用到的设计模式