Candies and Two Sisters

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辰chen 发表于 2022/06/14 22:49:48 2022/06/14
【摘要】 文章目录 一、Candies and Two Sisters总结 一、Candies and Two Sisters 本题链接:Candies and Two Sisters 题目:...


一、Candies and Two Sisters

本题链接:Candies and Two Sisters

题目:
A. Candies and Two Sisters
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
There are two sisters Alice and Betty. You have n candies. You want to distribute these n candies between two sisters in such a way that:

Alice will get a (a>0) candies;
Betty will get b (b>0) candies;
each sister will get some integer number of candies;
Alice will get a greater amount of candies than Betty (i.e. a>b);
all the candies will be given to one of two sisters (i.e. a+b=n).
Your task is to calculate the number of ways to distribute exactly n candies between sisters in a way described above. Candies are indistinguishable.

Formally, find the number of ways to represent n as the sum of n=a+b, where a and b are positive integers and a>b.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.

The only line of a test case contains one integer n (1≤n≤2⋅109) — the number of candies you have.

Output
For each test case, print the answer — the number of ways to distribute exactly n candies between two sisters in a way described in the problem statement. If there is no way to satisfy all the conditions, print 0.

Example
input
6
7
1
2
3
2000000000
763243547
output
3
0
0
1
999999999
381621773

Note
For the test case of the example, the 3 possible ways to distribute candies are:

a=6, b=1;
a=5, b=2;
a=4, b=3.

本博客给出本题截图:

题意:一共n个糖,给两个姐妹分糖吃,要求Alice得到的糖要比Betty多,问有几种分法

AC代码

#include <iostream>

using namespace std;

int main()
{
    int n;
    cin >> n;
    
    while (n -- )
    {
        int a;
        cin >> a;
        
        if (a % 2) cout << a / 2 << endl;
        else cout << (a - 1) / 2 << endl;
    }
    
    return 0;
}

  
 
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总结

观察输入输出猜规律

文章来源: chen-ac.blog.csdn.net,作者:辰chen,版权归原作者所有,如需转载,请联系作者。

原文链接:chen-ac.blog.csdn.net/article/details/117880089

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