Link: Educational Codeforces Round 72 (Rated for Div. 2)

Creating a Character

Description:
You play your favourite game yet another time. You chose the character you didn't play before. It has str points of strength and int points of intelligence. Also, at start, the character has exp free experience points you can invest either in strength or in intelligence (by investing one point you can either raise strength by 1 or raise intelligence by 1).

Since you'd like to make some fun you want to create a jock character, so it has more strength than intelligence points (resulting strength is strictly greater than the resulting intelligence).

Calculate the number of different character builds you can create (for the purpose of replayability) if you must invest all free points. Two character builds are different if their strength and/or intellect are different.

Input
The first line contains the single integer T (1≤T≤100) — the number of queries. Next T lines contain descriptions of queries — one per line.

This line contains three integers str, int and exp (1≤str,int≤108, 0≤exp≤1e8) — the initial strength and intelligence of the character and the number of free points, respectively.

Output
Print T integers — one per query. For each query print the number of different character builds you can create.

Example
input
4
5 3 4
2 1 0
3 5 5
4 10 6
output
3
1
2
0
Note
In the first query there are only three appropriate character builds: (str=7,int=5), (8,4) and (9,3). All other builds are either too smart or don't use all free points.

In the second query there is only one possible build: (2,1).

In the third query there are two appropriate builds: (7,6), (8,5).

In the fourth query all builds have too much brains.

Problem solving:
这道题的意思就是有一个角色，给了你他的智力值和体力值的大小，然后给了你现在可分配的点数。问你有多少种情况。可以在把所有点数分完之后体力值大于智力值。每个点数都可以分配给智力值或体力值而加一。
数据范围1e8*100，直接暴力肯定会超时。
所以我们可以用二分来写。找到能满足条件的最小的分配给体力的值，然后用总点数减去这个值就是可以达到目的的分配次数。二分中的check直接判断体力值加上mid与智力值加上总点数-mid的大小关系即可。
Code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a,b,e,x;
bool check(int x)
{
    if(a+x>(b+(e-x)))  return 1;
    return 0;
}
int main()
{
    ll n;
    cin>>n;
    while(n--)
    {
        cin>>a>>b>>e;
        ll l=0,r=e,mid;
        while(l<=r)
        {
            mid=(l+r)/2;
            if(check(mid))  r=mid-1;
            else    l=mid+1;
        }
        cout<<e-r<<endl;
    }
    return 0;
}
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