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Allowance
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 Hint
INPUT DETAILS:
FJ would like to pay Bessie 6 cents per week. He has 100 1-cent coins,120 5-cent coins, and 1 10-cent coin. OUTPUT DETAILS: FJ can overpay Bessie with the one 10-cent coin for 1 week, then pay Bessie two 5-cent coins for 10 weeks and then pay Bessie one 1-cent coin and one 5-cent coin for 100 weeks. Source |
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题意：给你n个面值的金币，每个金币有固定数量，每周你都要交至少m块钱，问你最多能撑多少周
题解：首先很容易想到的是先用大的，对于大于m面额的金币，直接答案加上该面值的数量。
对于小于m面值的金币，首先取大的，达到临界值后我往小的取，直到达到m，但是此时会出现两种情况。
（1）小的不管怎么加都加不到，呢我肯定还加大的。
（2）加超了，呢我就从前往后遍历，一个个的减去小的使得达到临界值即可。
一开始写超时了，于是我加了个优化：对于当前满足题意的组合方式，我开一个数组标记每个面值的金币用了几张
然后让这些面值的金币减去标记的最小非零值，然后就过了。。。。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
struct node
{
ll x,y;
}a[30];
ll b[30];
bool comp(node a,node b)
{
return a.x<b.x;
}
int main(void)
{
ll n,m,i,j,ans=0;
scanf("%lld%lld",&n,&m);
for(i=1;i<=n;i++)
scanf("%lld%lld",&a[i].x,&a[i].y);
sort(a+1,a+n+1,comp);
for(i=n;i>0;i--)
{
//printf("%lld\n",a[i].y);
if(a[i].y==0)
continue;
while(a[i].y!=0)
{
if(a[i].x>=m)
ans+=a[i].y,a[i].y=0;
else
{
ll tmp=m;
memset(b,0,sizeof(b));
while(a[i].y>0 && a[i].x<=tmp)
tmp-=a[i].x,a[i].y--,b[i]++;
for(j=i-1;j>0;j--)
{
while(a[j].y>0 && a[j].x<=tmp)
tmp-=a[j].x,a[j].y--,b[j]++;
}
if(tmp!=0)
{
for(j=1;j<=n;j++)
{
while(a[j].y>0 && tmp>0)
tmp-=a[j].x,a[j].y--,b[j]++;
}
}
if(tmp>0)
break;
else
{
ans++;
for(j=1;j<=n;j++)
{
if(b[j]==0)
continue;
while(b[j]>0 && tmp+a[j].x<=0)
tmp+=a[j].x,a[j].y++,b[j]--;
}
ll mins=1e18;
for(j=1;j<=n;j++)
{
if(b[j]==0)
continue;
mins=min(mins,a[j].y/b[j]);
}
//printf("%lld\n",mins);
for(j=1;j<=n;j++)
a[j].y-=b[j]*mins;
ans+=mins;
}
}
}
}
printf("%lld\n",ans);
return 0;
}
