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; }