zoukankan      html  css  js  c++  java
  • hdu6092 01背包

    Rikka with Subset

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 658    Accepted Submission(s): 297


    Problem Description
    As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

    Yuta has n positive A1An and their sum is m . Then for each subset S of A , Yuta calculates the sum of S .

    Now, Yuta has got 2n numbers between [0,m] . For each i[0,m] , he counts the number of i s he got as Bi .

    Yuta shows Rikka the array Bi and he wants Rikka to restore A1An .

    It is too difficult for Rikka. Can you help her?  
     
    Input
    The first line contains a number t(1t70) , the number of the testcases.

    For each testcase, the first line contains two numbers n,m(1n50,1m104) .

    The second line contains m+1 numbers B0Bm(0Bi2n) .
     
    Output
    For each testcase, print a single line with n numbers A1An .

    It is guaranteed that there exists at least one solution. And if there are different solutions, print the lexicographic minimum one.
     
    Sample Input
    2
    2 3
    1 1 1 1
    3 3
    1 3 3 1
     
    Sample Output
    1 2
    1 1 1
     
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int N=1e4+88;
    typedef long long LL;
    LL dp[N],B[N];
    int a[N];
    int main(){
        int n,m,T;
        for(scanf("%d",&T);T--;){
        memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));
        scanf("%d%d",&n,&m);
        for(int i=0;i<=m;++i) scanf("%I64d",&B[i]);
        dp[0]=1;
        for(int i=1;i<=m;++i){
            if(dp[i]==B[i]) continue;
            a[i]=B[i]-dp[i];
            for(int j=1;j<=a[i];++j) for(int k=m;k>=i;--k) dp[k]+=dp[k-i];
        }
        int i;
        for(i=1;i<=m;++i) if(a[i]--) {printf("%d",i);break;}
        for(;i<=m;++i) while(a[i]--) printf(" %d",i);
        puts("");
        }
    }
  • 相关阅读:
    strongswan--HA
    RFC8221 -- 密码算法实现要求和使用指南
    strongswan--ikev2软件架构
    strongswan--ike sa状态机
    strongswan--ipsec.conf配置文件分析
    strongswan--linux内核ipsec policy类型
    strongswan--配置Charon-systemd问题解决
    strongswan——IKE建立过程
    GCC入门
    MPLS-TP OAM各个层次
  • 原文地址:https://www.cnblogs.com/mfys/p/7308968.html
Copyright © 2011-2022 走看看