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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