sdut2408 Pick apples

Pick apples

Time Limit: 1000ms   Memory limit: 165536K  有疑问？点这里^_^

题目描述

Once ago, there is a mystery yard which only produces three kinds of apples. The number of each kind is infinite. A girl carrying a big bag comes into the yard. She is so surprised because she has never seen so many apples before. Each kind of apple has a size and a price to be sold. Now the little girl wants to gain more profits, but she does not know how. So she asks you for help, and tell she the most profits she can gain.

输入

In the first line there is an integer T (T <= 50), indicates the number of test cases.
In each case, there are four lines. In the first three lines, there are two integers S and P in each line, which indicates the size (1 <= S<= 100) and the price (1 <= P <= 10000) of this kind of apple.

In the fourth line there is an integer V，(1 <= V <= 100,000,000)indicates the volume of the girl's bag.

输出

For each case, first output the case number then follow the most profits she can gain.

示例输入

11 12 13 16

示例输出

Case 1: 6

提示

 

来源

2012年"浪潮杯"山东省第三届ACM大学生程序设计竞赛

大范围先贪心，小范围背包。

但背包的范围也不能太小，不然就不对了。

把10000该为100就会WA

贪心算法和背包算法要好好再看看

#include<iostream>#include<cstdio>#include<cstring>using namespace std;long long V,dp[20100],s[5],p[5];void CompletePack(long long ss, long long pp){    for(long long i=ss ; i<=V; i++)        if(dp[i] < dp[i - ss] + pp)        dp[i] = dp[i - ss] + pp;}int main(){    int t,cases=0;    scanf("%d", &t);    while(t--)    {        double val,max=0;        int id;        for(int i=0;i<3;i++)        {            cin>>s[i]>>p[i];            val = 1.0*p[i]/s[i];            if(max < val)            {                max=val;                id=i;            }        }        cin>>V;        memset(dp,0,sizeof(dp));        cout <<"Case "<<++cases <<": ";        if(V <= 10000){            for(int i=0; i<3; i++)                CompletePack(s[i], p[i]);            cout<<dp[V]<<endl;        }else{            long long ans = (V-10000) / s[id]*p[id];            V = V - (V - 10000) / s[id] * s[id];            for(int i=0; i<3; i++)                CompletePack(s[i], p[i]);            cout << dp[V] + ans <<endl;        }    }    return 0;}