牛客周赛 Round 140

第一题

一道模拟题，1分钟就想到思路了。直接在l到r这个区间内遍历一下，如果这个数和abc都不相等，cnt就++。最后输出就行了。

```#include <bits/stdc++.h>

#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sc scanf
#define pf printf
#define x first
#define y second

using namespace std;

typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> PII;

int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
int ddx[] = {-1, 1, 0, 0, -1, 1, -1, 1};
int ddy[] = {0, 0, -1, 1, -1, -1, 1, 1};

LL ksm(LL a, LL b, LL p)
{
    LL sum = 1;
    a %= p;
    while (b)
    {
        if (b & 1)  sum = (sum * a) % p;
        b >>= 1;
        a = (a * a) % p;
    }
    return sum % p;
}

bool is_prime(int x)
{
	if (x == 0 || x == 1)  return false;
	for (int i = 2; i * i <= x; i ++)
		if (x % i == 0)  return false;
	
	return true;
}

bool is_runyear(int y)
{
	return (y % 400 == 0 || (y % 100 != 0 && y % 4 == 0));
}

LL gcd(LL x, LL y)
{
	while (y)
	{
		x %= y;
		swap(x, y);
	}
	return x;
}

const int N = 110;

/*
    1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
    1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    1 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
    1 3 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    1 4 6 4 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    ...
*/

int a, b, c, l, r, cnt;

void solve()
{
    cin >> a >> b >> c >> l >> r;
    for (int i = l; i <= r; i ++)
    {
        if (i != a && i != b && i != c)
        {
            cnt ++;
        }
    }
    
    cout << cnt << endl;
}


int main()
{
	IOS;
	int _ = 1;
// 	cin >> _;
	while (_ --)
	{
		solve();
	}
	
	return 0;
}