[POJ3040] Allowance

最新推荐文章于 2023-07-09 16:03:37 发布

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最新推荐文章于 2023-07-09 16:03:37 发布

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原文链接：http://www.cnblogs.com/HLXZZ/p/7509446.html

本文介绍了一个关于计算给定硬币集合下，如何使用贪心算法确定支付Bessie至少固定金额报酬的最大可能周数的问题。通过将硬币按价值排序，并采取逐步策略实现目标支付。

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

题解：

贪心

贪心策略（分三步）：

1、显然先把大于c的全部出完

2、从大到小只要不超过c，能出就出

3、从小到大能出就出，直到刚好超过c

证明：显然每次出钱需要浪费的尽量少，这样每次出只会使得出的钱刚好等于c或大于c一点点，所以每次是最优的

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;

const int N =25;

struct Node {
  int a,b;
  bool operator < (const Node &x) const {
    return a<x.a;
  }
}p[N];

int gi() {
  int x=0,o=1; char ch=getchar();
  while(ch!='-' && (ch<'0' || ch>'9')) ch=getchar();
  if(ch=='-') o=-1,ch=getchar();
  while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return o*x;
}

int main() {
  int n=gi(),c=gi(),i,ans=0;
  for(i=1; i<=n; i++) p[i].a=gi(),p[i].b=gi();
  sort(p+1,p+n+1);
  i=n;
  while(p[i].a>c) ans+=p[i].b,i--;
  n=i;
  while(1) {
    int now=0;
    for(i=n; i>=1; i--) {
      while(p[i].b && now+p[i].a<=c) {
	now+=p[i].a;
	p[i].b--;
      }
    }
    for(i=1; i<=n; i++) {
      while(p[i].b && now<c) {
	now+=p[i].a;
	p[i].b--;
      }
    }
    if(now<c) break;
    ans++;
  }
  printf("%d", ans);
  return 0;
}

转载于:https://www.cnblogs.com/HLXZZ/p/7509446.html