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:
    英语语法入门十五(名词所有格)
    在线调试Arduino
    CANopen和Canfestival
    嵌入式系统中的printf
    云原生爱好者周刊:Lens 5.0 发布,更炫、更快、更强!
    基于 KubeSphere 的 Nebula Graph 多云架构管理实践
    KubeSphere Meetup 北京站火热报名中 | 搭载 CIC 2021 云计算峰会
    KubeSphere Helm 应用仓库源码分析
    开启 Calico eBPF 数据平面实践
    KubeSphere 在直播应用中的实践
  • 原文地址:https://www.cnblogs.com/crazyapple/p/3201979.html
Copyright © 2011-2022 走看看