Codeforces 812 C Sagheer and Nubian Market

题目链接：http://codeforces.com/contest/812/problem/C

题意：他给你n个数字，给你一个最多能花的代价，让你求最多能买多少的东西，只是他每个物品代价和要买多少个物品有关系（如买k件物品，则代价为 Axj + xj·k ）

题解：第一个想法就是二分，二分每次要买的东西个数，每次在按那个规则排序就能写出来了（PS居然是因为精度问题一直在wa，最后才发现）

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <map>#include <set>#include <algorithm>#define N 100010#define inf 0x3f3f3f3f#define LL long longusing namespace std;LL mid;LL cnt;LL n, m;struct node {    LL id, num;}a[N];bool cmp(node aa, node b) {    return aa.id*mid + aa.num < b.id*mid + b.num;}bool solve() {    LL sum = 0;    sort(a, a + n, cmp);    for (int i = 0; i < mid; i++) {        sum += a[i].id*mid + a[i].num;    }    if (sum <= m) {        cnt = sum;        return true;    }    return false;}int main() {    cin.sync_with_stdio(false);    while (cin >> n >> m) {        for (int i = 0; i < n; i++) {            cin >> a[i].num;            a[i].id = i + 1;        }        cnt = 0;        LL l = 0;        LL ans = 0;        LL r = n;        while (l <= r) {            mid = (l + r) >> 1;            if (solve()) {                ans = mid;                l = mid + 1;            }            else {                r = mid - 1;            }        }        cout << ans << " " << cnt << endl;    }    return 0;}