codeforces16B

最新推荐文章于 2025-03-28 23:10:50 发布

转载最新推荐文章于 2025-03-28 23:10:50 发布 · 125 阅读

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原文链接：http://www.cnblogs.com/gaojunonly1/p/10787440.html

本文详细解析了CodeForces平台上的经典算法问题'Burglar and Matches'，通过贪心算法策略，解决了一个关于如何在限定条件下最大化背包中物品总价值的问题。代码实现清晰，提供了完整的解决方案。

Burglar and Matches

CodeForces - 16B

A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are m containers, in the i-th container there are a_i matchboxes, and each matchbox contains b_i matches. All the matchboxes are of the same size. The burglar's rucksack can hold n matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than nmatchboxes so that the total amount of matches in them is maximal.

Input

The first line of the input contains integer n (1 ≤ n ≤ 2·10⁸) and integer m (1 ≤ m ≤ 20). The i + 1-th line contains a pair of numbers a_i and b_i (1 ≤ a_i ≤ 10⁸, 1 ≤ b_i ≤ 10). All the input numbers are integer.

Output

Output the only number — answer to the problem.

Examples

Input

Output

Input

Output

7

sol：显然取个数多的，然后直接按题意贪心模拟即可

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=100005;
int n,m;
struct Match
{
    int Ges,Cnt;
}Box[N];
inline bool cmp_Cnt(Match p,Match q)
{
    return p.Cnt>q.Cnt;
}
int main()
{
    int i,ans=0;
    R(n); R(m);
    for(i=1;i<=m;i++)
    {
        R(Box[i].Ges); R(Box[i].Cnt);
    }
    sort(Box+1,Box+m+1,cmp_Cnt);
    for(i=1;i<=m&&n;i++)
    {
        ans+=min(n,Box[i].Ges)*Box[i].Cnt;
        n-=min(n,Box[i].Ges);
    }
    Wl(ans);
    return 0;
}
/*
Input
7 3
5 10
2 5
3 6
Output
62

Input
3 3
1 3
2 2
3 1
Output
7
*/

View Code

转载于:https://www.cnblogs.com/gaojunonly1/p/10787440.html