Allowance
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5604 | Accepted: 2195 |
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
思路：
好题啊~
为了能提供更多的天数，我们肯定从面值大的钱开始（题中也说了大的面值是小面值的倍数，所以我们选大的，若换成小的则需要更多钱币）
从大面值开始一直凑到小于等于C，然后从小面值的凑够大于等于C，因为超出的钱越少，越好，所以从小面值开始。
还有一点要注意的：
比如C=20，现在大面值的凑到了16，小面值的有2、4，则选4不选两张2，也就是说，大面值凑的时候已经是最接近C的了，小面值应选1张让其满足就行了，若选两张2的话是有点浪费钱币张数了
代码如下：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
struct A{
int val,num;
}money[25];
int tmp[25];
bool cmp(A a,A b){
return a.val>b.val;
}
int main(){
int n,c,x,y;
scanf("%d%d",&n,&c);
int ans=0,m=0;
for(int i=0;i<n;i++){
scanf("%d%d",&x,&y);
if(x>=c)ans+=y;
else {
money[m].val=x;
money[m++].num=y;
}
}
sort(money,money+m,cmp);
//用tmp[i]数组临时记录，数i组成c，需要多少个
while(1){
memset(tmp,0,sizeof(tmp));
int s=c;
//这一轮凑够小于等于c需要的每个数多少
for(int i=0;i<m;i++){
tmp[i]=min(money[i].num,s/money[i].val);
s-=tmp[i]*money[i].val;
}
//在小于c的情况下，再从小的往大凑
if(s>0){
for(int i=m-1;i>=0;i--){
if(money[i].num>=1&&money[i].val>=s){
tmp[i]++;
s-=money[i].val;
break;
}
}
}
if(s>0)break;//如果怎么都凑不够一个c那就结束
int mymin=INF;
for(int i=0;i<m;i++){
if(tmp[i])mymin=min(mymin,money[i].num/tmp[i]);
}
ans+=mymin;
int flag=0;
for(int i=0;i<m;i++){
if(tmp[i]){
money[i].num-=(mymin*tmp[i]);
}
if(money[i].num)flag=1;
}
if(!flag)break;
}
printf("%d\n",ans);
}