poj 3040 Allowance

Allowance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1842		Accepted: 763

Description

As a reward for record milk production, Farmer John has decided to start paying Bessie the cow a small weekly allowance. FJ has a set of coins in N (1 <= N <= 20) different denominations, where each denomination of coin evenly divides the next-larger denomination (e.g., 1 cent coins, 5 cent coins, 10 cent coins, and 50 cent coins).Using the given set of coins, he would like to pay Bessie at least some given amount of money C (1 <= C <= 100,000,000) every week.Please help him ompute the maximum number of weeks he can pay Bessie.

Input

* Line 1: Two space-separated integers: N and C

* Lines 2..N+1: Each line corresponds to a denomination of coin and contains two integers: the value V (1 <= V <= 100,000,000) of the denomination, and the number of coins B (1 <= B <= 1,000,000) of this denomation in Farmer John's possession.

Output

* Line 1: A single integer that is the number of weeks Farmer John can pay Bessie at least C allowance

Sample Input

3 6
10 1
1 100
5 120

Sample Output

111

#include<algorithm>
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
struct TT
{
    int  v,w;
}a[25];
int need[25];
bool cmp(TT m, TT n)
{
  if(m.v>n.v) return true;
         return false;
}
int main()
{
    int n,i,k;
    int value,aa,bb,ans;
    while(~scanf("%d %d",&n,&value))
    {
        ans = 0;
        k = 0;
        for( i=0;i<n;i++)
        {
            scanf("%d %d",&aa,&bb);//排除大数；
            if( aa>= value)
            {
                 ans = ans+bb;
            }
            else
             {
                 a[k].v = aa;
                 a[k].w = bb;
                 k++;
             }
        }
        //k = n;
        //printf("sdfgsdf
");
        sort(a,a+k,cmp);
        while(1)
        {
            memset(need,0,sizeof(need));
            int sum = value;
            for(int i=0; i<k; i++) // Õý×ÅÕÒ
            {
                int tmp = sum/a[i].v;
                need[i] = min(a[i].w,tmp);
                sum = sum - a[i].v*need[i];
            }
            if(sum>0)
            {
                for(int i=k-1;i>=0;i--)
                {
                    if(a[i].w && a[i].v>=sum)
                    {
                        need[i]++;
                        sum = 0;
                        break;
                    }
                }
            }
            if(sum>0) break;
            int s = 0x3f3f3f3f;
            for(int i=0;i<k;i++)
            {
                if(need[i])
                    s = min(s,a[i].w/need[i]);
            }
            ans = ans+s;
            for(int i=0;i<k;i++)
            {
                if(need[i])
                    a[i].w -= need[i]*s;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}