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Language:
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 |
思路：可以贪心（因为面值之间是相互整除的），面值比c大的一天发一张，比c小的就优先选面值大的，从大到小贪一遍，这个过程中要保证所选硬币面值之和m要小于等于c，若m小于c，则从小往大优先选一张能使m大于等于c，这样是最优的。（开始没用need，1000ms险过。。。后来看到网上用的need，0ms。。。太菜）
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
struct Money
{
int v,b;
}cent[21];
int n,c;
int need[21];
int cmp(Money x,Money y)
{
return x.v<y.v;
}
int main()
{
int i,j;
while (~sff(n,c))
{
FRL(i,0,n)
sff(cent[i].v,cent[i].b);
sort(cent,cent+n,cmp);
int r=n-1;
int ans=0;
while (cent[r].v>=c){ //比c大的直接加上数量，一天发一张
ans+=cent[r].b;
cent[r].b=0;
r--;
}
while (1)
{
mem(need,0);
int mm=c;
for (i=r;i>=0;i--) //从大到小
{
int num=min(cent[i].b , mm/cent[i].v);
mm-=num*cent[i].v;
need[i]=num;
}
if (mm>0) //没凑够的从小往大的找一张能凑够的最小的面值
{
FRL(i,0,n)
{
if (cent[i].b>0&&mm-cent[i].v<=0){
need[i]++;
mm-=cent[i].v;
break;
}
}
}
if (mm>0) break;
int minn=INF;
FRL(i,0,n)
if (need[i])
minn=min(minn,cent[i].b/need[i]);
ans+=minn;
FRL(i,0,n)
if (need[i])
cent[i].b-=minn*need[i];
}
printf("%d\n",ans);
}
return 0;
}
原文地址：http://blog.csdn.net/u014422052/article/details/44114827
