题目链接
https://vjudge.net/problem/CodeForces-1121C
题面
Description
Vasya likes taking part in Codeforces contests. When a round is over, Vasya follows all submissions in the system testing tab.
There are (n) solutions, the (i)-th of them should be tested on (a_i) tests, testing one solution on one test takes (1) second. The solutions are judged in the order from (1) to (n). There are (k) testing processes which test solutions simultaneously. Each of them can test at most one solution at a time.
At any time moment (t) when some testing process is not judging any solution, it takes the first solution from the queue and tests it on each test in increasing order of the test ids. Let this solution have id (i), then it is being tested on the first test from time moment (t) till time moment (t + 1), then on the second test till time moment (t + 2) and so on. This solution is fully tested at time moment (t + a_i), and after that the testing process immediately starts testing another solution.
Consider some time moment, let there be exactly (m) fully tested solutions by this moment. There is a caption "System testing: (d)%" on the page with solutions, where (d) is calculated as
where (round(x) = lfloor{x + 0.5} floor) is a function which maps every real to the nearest integer.
Vasya calls a submission interesting if there is a time moment (possibly, non-integer) when the solution is being tested on some test (q), and the caption says "System testing: (q)%". Find the number of interesting solutions.
Please note that in case when multiple processes attempt to take the first submission from the queue at the same moment (for instance, at the initial moment), the order they take the solutions does not matter.
Input
The first line contains two positive integers (n) and (k) ((1 le n le 1000), (1 le k le 100)) standing for the number of submissions and the number of testing processes respectively.
The second line contains (n) positive integers (a_1, a_2, ldots, a_n) ((1 le a_i le 150)), where (a_i) is equal to the number of tests the (i)-th submission is to be run on.
Output
Output the only integer — the number of interesting submissions.
Examples
Input
2 2
49 100
Output
1
Input
4 2
32 100 33 1
Output
2
Input
14 5
48 19 6 9 50 20 3 42 38 43 36 21 44 6
Output
5
Note
Consider the first example. At time moment (0) both solutions start testing. At time moment (49) the first solution is fully tested, so at time moment (49.5) the second solution is being tested on the test (50), and the caption says "System testing: (50)%" (because there is one fully tested solution out of two). So, the second solution is interesting.
Consider the second example. At time moment (0) the first and the second solutions start testing. At time moment (32) the first solution is fully tested, the third solution starts testing, the caption says "System testing: (25)%". At time moment (32 + 24.5 = 56.5) the third solutions is being tested on test (25), the caption is still the same, thus this solution is interesting. After that the third solution is fully tested at time moment (32 + 33 = 65), the fourth solution is fully tested at time moment (65 + 1 = 66). The captions becomes "System testing: (75)%", and at time moment (74.5) the second solution is being tested on test (75). So, this solution is also interesting. Overall, there are two interesting solutions.
题意
给定(n)个题目，每个题目有(a[i])个测试点，有k台评测机，每台评测机在某一时刻只能测试一个题目，把所有测试点测完后测试队列中的下一道题，测试的进度用(d = roundleft(100cdotfrac{m}{n} ight))来表示，其中(round(x) = lfloor{x + 0.5} floor) ，m是完全测完的题目数量，对于每一个测试中的题，如果某一时刻正好测试到(i_{th})测试点，而(i_{th}=d)则这个题目被称为有趣的，问有多少有趣的题目
题解
我们可以用一个优先队列很快的算出每个题测试的开始时间和结束时间，然后我们就可以知道每个时刻前有多少个题目已经测试完毕了，所以就能算出每个时刻的测试进度，以及每个进度对应的时间区间，然后我们对于每一个题，遍历每一个时间进度覆盖的区间，如果和当前题结束和开始的时间段有交集就判断是否有趣。
至于如何判断，如果当前进度覆盖的区间刚好能包括进去当前题目的测试对应等于进度测试点的时间，那么就是有趣的，这个时间就是这个题目开始的时间+这个区间对应的进度值,，这个时间要大于等于这个进度的左端点，同时要小于等于这个题目测试完的时间和这个区间右端点的较小值。
AC代码
#include <bits/stdc++.h>
#define N 1050
using namespace std;
int a[N];
int min(int a, int b) {
return a < b ? a : b;
}
struct node {
int id, val;
node (int id = 0, int val = 0): id(id), val(val) {}
bool operator < (const node &b) const {
return val > b.val;
}
};
struct fin {
int l, r;
} finish[N];
int pre[150 * N];
int round1[150 * N];
struct pro{
int l; int r; int val;
} process[N];
priority_queue<node> q;
int main() {
int n, k;
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
for (int i = 1; i <= n; i++) {
if (q.size() < k) q.push(node(i, a[i]));
else {
node now = q.top();
finish[now.id].l = now.val - a[now.id];
finish[now.id].r = now.val;
q.pop();
q.push(node(i, a[i] + now.val));
}
}
int last = 0;
while (!q.empty()) {
node now = q.top();
finish[now.id].l = now.val - a[now.id];
finish[now.id].r = now.val;
last = max(last, now.val);
q.pop();
}
// for (int i = 1; i <= n; i++) {
// cout << finish[i].l << " " << finish[i].r << endl;
// }
for (int i = 1; i <= n; i++) {
pre[finish[i].r]++;
}
for (int i = 1; i <= last; i++) {
pre[i] += pre[i - 1];
}
for (int i = 1; i <= last; i++) {
round1[i] = floor((double)pre[i] / (double)n * 100 + 0.5);
}
int cnt = 0;
int tmp = 0;
for (int i = 1; i <= last; i++) {
if (round1[i] != round1[i - 1]) {
process[++cnt].val = round1[i];
process[cnt].l = i + 1;
if (cnt != 0) {
process[cnt - 1].r = i;
}
}
}
process[cnt].r = last + 1;
// for (int i = 1; i <= cnt; i++) {
// cout << process[i].l << " " << process[i].r << " " << process[i].val << endl;
// }
int ans = 0;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= cnt; j++) {
if (process[j].l <= finish[i].r && process[j].r >= finish[i].l) {
if (process[j].l <= finish[i].l + process[j].val && min(finish[i].r, process[j].r) >= finish[i].l + process[j].val) {
// cout << "ans: " << i << endl;
ans++;
break;
}
}
}
}
printf("%d
", ans);
return 0;
}