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[];
 int need[];
 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 = ;
         k = ;
         for( i=;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\n");
         sort(a,a+k,cmp);
         while()
         {
             memset(need,,sizeof(need));
             int sum = value;
             for(int i=; 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>)
             {
                 for(int i=k-;i>=;i--)
                 {
                     if(a[i].w && a[i].v>=sum)
                     {
                         need[i]++;
                         sum = ;
                         break;
                     }
                 }
             }
             if(sum>) break;
             int s = 0x3f3f3f3f;
             for(int i=;i<k;i++)
             {
                 if(need[i])
                     s = min(s,a[i].w/need[i]);
             }
             ans = ans+s;
             for(int i=;i<k;i++)
             {
                 if(need[i])
                     a[i].w -= need[i]*s;
             }
         }
         printf("%d\n",ans);
     }
     return ;
 }