1685: [Usaco2005 Oct]Allowance 津贴

1685: [Usaco2005 Oct]Allowance 津贴

Time Limit: 5 Sec  Memory Limit:
64 MB
Submit: 193  Solved: 141

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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 compute
the maximum number of weeks he can pay Bessie.

作为对勤勤恳恳工作的贝茜的奖励，约翰已经决定开始支付贝茜一个小的每周津贴．  约翰有n(1≤N≤20)种币值的硬币，面值小的硬币总能整除面值较大的硬币．比如说，币值有如下几种：1美分，5美分，10美分，50美分…．．
    利用给定的这些硬币，他将要每周付给贝茜一定金额的津贴C(1≤C≤10^8)．
    请帮他计算出他最多能给贝茜发几周的津贴．

Input

    第1行：2个用空格隔开的整数n和C.
    第2到n+1行：每行两个整数表示一种币值的硬币．第一个整数V(I≤y≤10^8)，表示币值．
第二个整数B(1≤B≤10^6)，表示约翰拥有的这种硬币的个数．

Output

    一个整数，表示约翰付给贝茜津贴得最多的周数．

Sample Input

3 6

10 1

1 1 00

5 1 20

Sample Output

111

样例说明

约翰想要每周付给贝茜6美分．他有1个10美分的硬币、100个1美分的硬币、120个5美分的硬币．约翰可以第一周付给贝茜一个10美分的硬币，接着的10周每周付给贝茜2个5芙分硬币，接下来的100周每周付给贝茜一个1美分的硬币和1个5美分的硬币．共计111周．

HINT

Source

 

对这题显然每周花的冤枉钱越少越好。。

因为前面的硬币总能整除后面的，所以排个序从大到小去找能找到最优方案。。。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxn = 30;

struct C{
int v,b;
bool operator < (const C &B) const {
return v < B.v;
}
}coin[maxn];

int n,i,j,C,h[maxn];
long long ans = 0;

int main()
{
#ifdef YZY
freopen("yzy.txt","r",stdin);
#endif

cin >> n >> C;
for (i = 1; i <= n; i++) scanf("%d%d",&coin[i].v,&coin[i].b);
sort (coin + 1,coin + n + 1);
while (1)
{
int rest = C;
memset(h,0,sizeof(h));
for (i = n; i > 0; i--)
{
int temp = min(coin[i].b,rest / coin[i].v);
rest -= temp * coin[i].v;
h[i] = temp;
}
if (rest > 0)
for (i = 1; i <= n; i++)
{
if (rest <= 0) break;
if (coin[i].b > h[i])
{
++h[i];
rest -= coin[i].v;
}
}
if (rest <= 0)
{
int tot = 1E9;
for (i = 1; i <= n; i++)
if (h[i] > 0) tot = min(tot,coin[i].b / h[i]);
ans += 1LL*tot;
for (i = 1; i <= n; i++)
if (h[i] > 0) coin[i].b -= tot * h[i];
}
else break;
}
cout << ans;
return 0;
}