zoukankan      html  css  js  c++  java
  • HDU 1171 Big Event in HDU (母函数 | 背包 )

    Big Event in HDU

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16262    Accepted Submission(s): 5741

    Problem Description
    Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002. The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).
     
    Input
    Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different. A test case starting with a negative integer terminates input and this test case is not to be processed.
     
    Output
    For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.
     
    Sample Input
    2
    10 1
    20 1
    3
    10 1
    20 2
    30 1
    -1
     
    Sample Output
    20 10
    40 40
     
    Author
    lcy
     
     
     1,母函数:
     
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    
    using namespace std;
    
    const int N=250010;
    
    int n,val[60],amount[60];
    int c1[N],c2[N];
    
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        while(~scanf("%d",&n) && n>0){      //天杀的n>0,我写成 n!=-1 WA了N久
            int tot=0;
            for(int i=0;i<N;i++){
                c1[i]=0;
                c2[i]=0;
            }
            for(int i=1;i<=n;i++){
                scanf("%d%d",&val[i],&amount[i]);
                tot+=val[i]*amount[i];
            }
            c1[0]=1;
            int maxx=0;
            for(int i=1;i<=n;i++){
                maxx+=val[i]*amount[i];
                for(int j=0;j<=maxx;j++)
                    for(int k=0;k<=amount[i] && j+k*val[i]<=maxx;k++)
                        c2[j+k*val[i]]+=c1[j];
                for(int j=0;j<=maxx;j++){
                    c1[j]=c2[j];
                    c2[j]=0;
                }
            }
            for(int i=tot/2;i>=0;i--)
                if(c1[i]!=0){
                    printf("%d %d\n",tot-i,i);
                    break;
                }
        }
        return 0;
    }

    2,背包:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    
    using namespace std;
    
    const int N=250010;
    
    int n,val[60],amount[60];
    int dp[N];
    
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        while(~scanf("%d",&n) && n>0){      //天杀的n>0,我写成 n!=-1 WA了N久
            int tot=0;
            for(int i=0;i<n;i++){
                scanf("%d%d",&val[i],&amount[i]);
                tot+=val[i]*amount[i];
            }
            memset(dp,0,sizeof(dp));
            dp[0]=1;
            for(int i=0;i<n;i++)
                for(int j=0;j<amount[i];j++)
                    for(int k=tot/2;k>=val[i];k--)
                        dp[k]+=dp[k-val[i]];
            int i;
            for(i=tot/2;i>=0;i--)
                if(dp[i]!=0){
                    printf("%d %d\n",tot-i,i);
                    break;
                }
            if(i==-1){  //如果到最后还不能分,则输出以下结果
                printf("%d %d\n",tot,0);
            }
        }
        return 0;
    }
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    
    using namespace std;
    
    const int N=256010;
    
    int n,val[60],amount[60];
    int dp[N];
    
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        while(~scanf("%d",&n) && n>0){
            int tot=0;
            for(int i=0;i<n;i++){
                scanf("%d%d",&val[i],&amount[i]);
                tot+=val[i]*amount[i];
            }
            memset(dp,0,sizeof(dp));
            for(int i=0;i<n;i++)
                for(int j=0;j<amount[i];j++)
                    for(int k=tot/2;k>=val[i];k--)
                        dp[k]=max(dp[k],dp[k-val[i]]+val[i]);
            printf("%d %d\n",tot-dp[tot/2],dp[tot/2]);
        }
        return 0;
    }
  • 相关阅读:
    张艾迪:全面开放“A"计划
    张艾迪(创始人):我们是激情与开放的世界
    张艾迪(创始人):由产品集合到创新的引领
    张艾迪(创始人):Be.Time+Cr.Idear的创新理念
    张艾迪(创始人): 整合全新的UIW.AD概念
    张艾迪(创始人): 整合全新的UIW.AD概念模式
    腾讯会员免费领取试用免费腾讯视频vip共享
    golang二进制bit位的常用操作
    golang ntp协议客户端
    imagecreatefrombmp、imagebmp php处理bmp文件
  • 原文地址:https://www.cnblogs.com/jackge/p/3028207.html
Copyright © 2011-2022 走看看