Mancunian Hoards Black Money，贪心，思维

 Mancunian Hoards Black Money - Problems - CodeChef

Mancunian is a real estate dealer in Mancunia. He is part of The Elite Builders Society who work hard day and night to improve the infrastructure of thecity, but they need cash to do this. They have a unique way to increase the amount of black money they possess. They start with a pile of cash amounting to S dollars. Then they repeat the following process N times. Before the ith step, there are already i piles of cash on the table. They create another pile amounting to the sum of amounts of all the i previous piles and add another Ci dollars to it.

A hot piece of property priced at X dollars has just appeared on the market and the society decided that they must buy it at all costs. They need to select a few of the piles from the ones on the table, whose sum is exactly X.

Although shrewd businessmen, they lack mathematical skills. Can you help them decide whether it is possible or not?

Input

The first line of the file contains an integer T denoting the number of test cases.

Each case consists of two lines.

The first line contains three integers S, N and X denoting the amount of cash in the first pile, the number of times the society creates a new pile and the cost of the in-demand property respectively.

The second line contains N integers denoting the values Ci mentioned in the problem description.

Output

Output T lines. In each line, output "yes" if the society can buy the property and "no" otherwise.

Constraints

• 1 ≤ T, N ≤ 105
• 1 ≤ sum of N over all test cases ≤ 105
• 1 ≤ S, X, Ci ≤ 1018

Example

Input:
2
1 4 6
1 2 4 2
100 2 500
51 749

Output:
yes
no

Explanation

Example case 1: The piles of cash are 1, 2, 5, 12 and 22. The society will use the piles of cash amounting to 1 and 5 to get a total of 6.

Example case 2: The piles of cash are 100, 151 and 1000. The society cannot buy the property costing 500 using any of the piles of cash they have.

解析：

贪心；
我们需要根据题目所给的 S 和 c[i] 推出每堆 cash 的值 a[i] ，在此过程中我们还会推出 a[i] 的前缀和数组 sum[[i];
通过观察我们发现如果 X > sum[i] 那么 X 一定加数中一定有 a[i+1];
因为:

sum[i+1]>X>sum[i],如果没有 a[i+1]，那么 sum[i] 中一定没有能够凑到 X 的数，且通过a序列的生成方式我们知道 a[i+2] 一定大于 sum[i] ,所以当 sum[i+1]>X ，X 的加数中一定不可能存在 a[i+1] 后的数。

让我们举一个例子。对于给定的${C_i} 值，设级数为${1,3,6,15,30,70}$。让我们看看值为30的桩。在此之前的所有桩的总和为1+3+6+15=25，小于30。
现在，如果$X$的值在（30,70）范围内，你会怎么做？注意，如果我们去掉堆30，我们就不能形成30到70之间的任何值。如果我们排除桩30，则桩30之前的桩不能用于形成桩30和70之间的值。这是因为即使包括30之前的所有桩，我们也无法形成超过30的值。我们完全依赖于桩30来形成30-70之间的值。
 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
const int N = 1e5 + 5;
LL S ,n,X;
LL c[N],a[N], sum[N];


int main() {
	int T;
	cin >> T;
	while (T--) {
		scanf("%lld%lld%lld", &S, &n, &X);
		for (int i = 1; i <= n; i++) {
			scanf("%lld", &c[i]);
		}
		a[1] = S;
		sum[1] = S;
		int cnt = 1;
		for (int i = 2; i <= n+1; i++) {
			if (sum[i - 1] + c[i - 1] > 1e18)
				break;
			a[i] = sum[i - 1] + c[i - 1];
			sum[i] = sum[i - 1] + a[i];
			cnt++;
		}
		for (int i = cnt; i >= 0; i--) {
			if (X>sum[i]) {
				X -= a[i+1];
			}
		}
		if (X == 0)
			cout << "yes" << endl;
		else
			cout << "no" << endl;
	}
	return 0;
}