A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are mcontainers, in the i-th container there are ai matchboxes, and each matchbox contains bi 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 n matchboxes so that the total amount of matches in them is maximal.
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 ai and bi (1 ≤ ai ≤ 108, 1 ≤ bi ≤ 10). All the input numbers are integer.
Output the only number — answer to the problem.
7 3 5 10 2 5 3 6
62
3 3 1 3 2 2 3 1
7
非常水的贪心 直接每次选装的多的盒子 n减去盒子数即可了
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int M = 25;
int a[M], b[M], c[M], n, m, ans;
bool cmp (int i, int j)
{
return b[i] > b[j];
}
int main()
{
scanf ("%d%d", &n, &m);
for (int i = 1; i <= m; ++i)
{
c[i] = i;
scanf ("%d%d", &a[i], &b[i]);
}
sort (c + 1, c + m + 1, cmp);
ans = 0;
for (int i = 1; i <= m; ++i)
{
if (n >= a[c[i]])
{
n -= a[c[i]];
ans += a[c[i]] * b[c[i]];
}
else
{
ans += n * b[c[i]];
break;
}
}
printf ("%d\n", ans);
return 0;
}
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