技術標籤：貪心演算法

題目：
http://codeforces.com/contest/1476/problem/B
You have a statistic of price changes for one product represented as an array of n positive integers p0,p1,…,pn−1, where p0 is the initial price of the product and pi is how the price was increased during the i-th month.

Using these price changes you are asked to calculate the inflation coefficients for each month as the ratio of current price increase pi to the price at the start of this month (p0+p1+⋯+pi−1).

Your boss said you clearly that the inflation coefficients must not exceed k%, so you decided to increase some values pi in such a way, that all pi remain integers and the inflation coefficients for each month don’t exceed k%.

You know, that the bigger changes — the more obvious cheating. That’s why you need to minimize the total sum of changes.

What’s the minimum total sum of changes you need to make all inflation coefficients not more than k%?

Input
The first line contains a single integer t(1≤t≤1000) — the number of test cases.
The first line of each test case contains two integers n and k (2≤n≤100; 1≤k≤100) — the length of array p and coefficient k.

Output
For each test case, print the minimum total sum of changes you need to make all inflation coefficients not more than k%.

Input

2
4 1
20100 1 202 202
3 100
1 1 1

Output

99
0

Note

In the first test case, you can, for example, increase p0 by 50 and p1 by 49 and get array [20150,50,202,202]. Then you get the next inflation coefficients:

In the second test case, you don’t need to modify array p, since the inflation coefficients are already good:

思路：
記錄字首和pre[]，如果我的當月的a[i] / pre[i-1] > k / 100，也就是a[i] * 100 > pre[i-1] * k，只有這種情況可以增加一些值pi。
根據貪心策略，我的每次的要增加只在第一個月增加而不是在當前的月增加，使得a[i] * 100 < (pre[i-1] + pi) * k 成立。根據貪心策略，每次的增加只在第一個月增加，那麼當月的前面所有月都滿足a[i] / pre[i-1] <= k / 100（分母增加只會使這個數變小）。然後逐個月取所需要的最大增加的pi。

程式碼：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN = 110;
ll a[MAXN], pre[MAXN];
int k, n;
ll sum;
int main(){
//    freopen("_in.txt", "r", stdin);
//    freopen("_out1.txt", "w", stdout);
	int t;
	 scanf("%d", &t);
	 while (t--){
	 	scanf("%d%d", &n, &k);
	 	pre[0] = 0;
	 	for (int i = 1; i <= n; i++){
	 		scanf("%lld", &a[i]);
			pre[i] = pre[i-1] + a[i];
		 }
//	 	if (k == 100){  //k = 100 時增加量不一定為0，有可能下一個月比前面所有月的總和的數要大
//	 		printf("0\n");
//	 		continue;
//		}
		sum = 0;
		for (int i = 2; i <= n; i++){
			if (a[i] * 100 > pre[i-1] * k){
				ll temp = a[i] * 100 / k;
					if (a[i] * 100 % k != 0)  //向上取整
						temp++;
				sum = max(sum, temp - pre[i-1]);
			}
		}
		printf("%lld\n", sum);
	 }
    return 0;
}