Codeforces Round #750 (Div. 2) - B. Luntik and Subsequences - 题解

Posted 2021-11-22 Tisfy

目录

• • Codeforces Round #750 (Div. 2) - B. Luntik and Subsequences
  • • Problem Description
    • Input
    • Sample Input
    • Sample Onput
    • Note
• 题目大意
• 解题思路
• AC代码

Codeforces Round #750 (Div. 2) - B. Luntik and Subsequences

传送门
Time Limit: 1 seconds
Memory Limit: 256 megabytes

Problem Description

Luntik came out for a morning stroll and found an array a of length $n$. He calculated the sum s of the elements of the array $(s={\sum }_{i=1}^n{a}_i$). Luntik calls $a$ subsequence of the array a nearly full if the sum of the numbers in that subsequence is equal to $s-1$.

Luntik really wants to know the number of nearly full subsequences of the array $a$. But he needs to come home so he asks you to solve that problem!

A sequence $x$ is a subsequence of a sequence $y$ if $x$ can be obtained from $y$ by deletion of several (possibly, zero or all) elements.

Input

The first line contains a single integer $t(1\le t\le 1000)$ — the number of test cases. The next $2\cdot t$ lines contain descriptions of test cases. The description of each test case consists of two lines.

The first line of each test case contains a single integer $n(1\le n\le 60)$ — the length of the array.

The second line contains $n$ integers $a_1,a_2,\dots ,a_n(0\le a_i\le 10^9)$ — the elements of the array $a$.

Sample Input

5
5
1 2 3 4 5
2
1000 1000
2
1 0
5
3 0 2 1 1
5
2 1 0 3 0

Sample Onput

1
0
2
4
4

Note

In the first test case, $s=1+2+3+4+5=15$, only ($2,3,4,5$) is a nearly full subsequence among all subsequences, the sum in it is equal to $2+3+4+5=14=15-1$.

In the second test case, there are no nearly full subsequences.

In the third test case, $s=1+0=1$, the nearly full subsequences are ($0$) and () (the sum of an empty subsequence is $0$).

---

题目大意

给你一个长度为$n$的序列，问你它有多少个 和为序列的和-1的子序列。

解题思路

知道原始序列后，我们可以立即求出原始序列的和（记为$s$），因此我们只需要找和为$s-1$的子序列（即为原始序列去掉一个$1$和若干个$0$）

我们可以统计出原始序列共有多少个$1$（记为$num1$）

对于每一个$1$，去掉这个$1$的话，剩下的元素还能去掉若干个$0$

假如共有$num0$个$0$，那么去掉若干个$0$的方案数是$2^{num0}$
也就是说，不论去掉哪一个$1$，都有$2^{num0}$个去掉$0$的方法，使得剩下的子序列的和为$s-1$

因为原始共有$num1$个$1$，根据乘法原理，去掉$1$个$1$和若干个$0$的方案数为$num1\times 2^{num0}$

所以直接输出$num1\times 2^{num0}$即可。

---

AC代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;
ll Pow[64] = {1}; // Pow[i]是2的i次方
void init() {
    for (int i = 1; i < 63; i++) {
        Pow[i] = Pow[i - 1] * 2;
    }
}
int a[100];
int main() {
    init();
    int N;
    cin >> N; // 共N组样例
    while(N--) {
        int n;
        cd(n); // scanf("%d", &n); 这组有n个元素
        int num1 = 0; // 1的个数
        int num0 = 0; // 0的个数
        fi(i, 0, n) { // for (int i = 0; i < n; i++)
            cd(a[i]); // scanf("%d", &a[i]);
            if (a[i] == 1) num1++;
            if (a[i] == 0) num0++;
        }
        ll ans = Pow[num0] * num1; // num1 * 2 ^ num0
        printf("%lld\\n", ans);
    }
    return 0;
}

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Tisfy：https://letmefly.blog.csdn.net/article/details/120939771