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）

 1 #include <iostream>
 2 #include <cstring>
 3 #include <string>
 4 #include <vector>
 5 #include <queue>
 6 #include <map>
 7 #include <set>
 8 #include <algorithm>
 9 #define N 100010
10 #define inf 0x3f3f3f3f
11 #define LL long long
12 using namespace std;
13 LL mid;
14 LL cnt;
15 LL n, m;
16 struct node {
17     LL id, num;
18 }a[N];
19 bool cmp(node aa, node b) {
20     return aa.id*mid + aa.num < b.id*mid + b.num;
21 }
22 bool solve() {
23     LL sum = 0;
24     sort(a, a + n, cmp);
25     for (int i = 0; i < mid; i++) {
26         sum += a[i].id*mid + a[i].num;
27     }
28     if (sum <= m) {
29         cnt = sum;
30         return true;
31     }
32     return false;
33 }
34 int main() {
35     cin.sync_with_stdio(false);
36     while (cin >> n >> m) {
37         for (int i = 0; i < n; i++) {
38             cin >> a[i].num;
39             a[i].id = i + 1;
40         }
41         cnt = 0;
42         LL l = 0;
43         LL ans = 0;
44         LL r = n;
45         while (l <= r) {
46             mid = (l + r) >> 1;
47             if (solve()) {
48                 ans = mid;
49                 l = mid + 1;
50             }
51             else {
52                 r = mid - 1;
53             }
54         }
55         cout << ans << " " << cnt << endl;
56     }
57     return 0;
58 }

2017-06-02 09:21:38

posted @ 2017-06-02 09:22  ガ落涙『不變』  阅读(87)  评论(0)  编辑  收藏  举报