zoukankan      html  css  js  c++  java
  • noj [1479] How many (01背包||DP||DFS)

    • http://ac.nbutoj.com/Problem/view.xhtml?id=1479
    • [1479] How many

    • 时间限制: 1000 ms 内存限制: 65535 K
    • 问题描述
    • There are N numbers, no repeat. All numbers is between 1 and 120, and N is no more than 60. then given a number K(1 <= K <= 100). Your task is to find out some given numbers which sum equals to K, and just tell me how many answers totally,it also means find out hwo many combinations at most.
    • 输入
    • There are T test cases. For each case: First line there is a number N, means there are N numbers. Second line there is a number K, means sum is K. Third line there lists N numbers. See range details in describe.
    • 输出
    • Output the number of combinations in total.
    • 样例输入
    • 2
      5
      4
      1,2,3,4,5
      5
      6
      1,2,3,4,5
      
    • 样例输出
    • 2
      3
    • 01背包||DP代码:
    •  1 #include <iostream>
       2 #include <stdio.h>
       3 #include <math.h>
       4 #include <string.h>
       5 
       6 using namespace std;
       7 int dp[110];
       8 
       9 int main()
      10 {
      11     int t;
      12     scanf("%d",&t);
      13     while(t--)
      14     {
      15         int n,m;
      16         scanf("%d%d",&n,&m);
      17         memset(dp,0,sizeof(dp));
      18         dp[0]=1;
      19         int i,j;
      20         for(i=0;i<n;i++)
      21         {
      22             int a;
      23             scanf("%d",&a);
      24             getchar();
      25             for(j=m;j>=0;j--)
      26             {
      27                 if(a+j<=m)
      28                 {
      29                     dp[a+j]+=dp[j];
      30                 }
      31             }
      32         //    for(j=0;j<=m;j++) printf("%d ",dp[j]); putchar(10);
      33         }
      34         printf("%d
      ",dp[m]);
      35     }
      36     return 0;
      37 }

      DFS代码:

       1 #include <iostream>
       2 #include <stdio.h>
       3 #include <string.h>
       4 
       5 using namespace std;
       6 
       7 int n,m;
       8 int a[100],mark[100];
       9 int cnt;
      10 
      11 void dfs(int x,int s){
      12     if(s==0){
      13         cnt++;
      14         return;
      15     }
      16     int i;
      17     for(i=x;i<n;i++){
      18         if(!mark[i]&&s-a[i]>=0){
      19             mark[i]=1;
      20             dfs(i+1,s-a[i]);
      21             mark[i]=0;
      22         }
      23     }
      24 }
      25 
      26 int main()
      27 {
      28     int t;
      29     scanf("%d",&t);
      30     while(t--){
      31         scanf("%d%d",&n,&m);
      32         int i;
      33         for(i=0;i<n;i++){
      34             scanf("%d",&a[i]);
      35             getchar();
      36         }
      37         memset(mark,0,sizeof(mark));
      38         cnt=0;
      39         dfs(0,m);
      40         printf("%d
      ",cnt);
      41     }
      42     return 0;
      43 }
  • 相关阅读:
    JS Ajax跨域访问
    CentOS 6.8 Java 环境搭建
    vue+vant ui+高德地图的选址组件
    vue和element全局loading
    axios简单的二次封装
    vuex的简单教程
    vue 使用 element ui动态添加表单
    Promise对象和async函数
    css不定高图文垂直居中的三种方法
    js点击复制文本
  • 原文地址:https://www.cnblogs.com/crazyapple/p/3201979.html
Copyright © 2011-2022 走看看