“浪潮杯”第九届山东省ACM大学生程序设计竞赛 F-Four-tuples (容斥)

链接：https://www.nowcoder.com/acm/contest/123/F
来源：牛客网

题目描述

Given    ,please count the number of four-tuples  such that and   

The answer should modulo10 ⁹+7 before output.

输入描述:

The input consists of several test cases. The first
line gives the number of test cases,T(1≤T≤106).

For each test case, the input contains one line
with 8 integers,.

输出描述:

For each test case, output one line containing one
integer, representing the answer.

1
1 1 2 2 3 3 4 4

1

题意：

给定四个区间(li,ri)(li,ri)（闭区间）,求一个四元组(x1,x2,x3,x4)(x1,x2,x3,x4)，满足xixi在区间(li,ri)(li,ri)内，且任意两个相邻的xixi不能相等。

题解：

容斥原理。就是麻烦。 

ans=所有都不相等-两个相等+三个相等+两对两个相等-四个相等。

A=（区间1与区间2的共有长度）*（区间3的长度）*（区间4的长度）

别人的代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const long long mod = 1e9 + 7;
long long l1,l2,l3,l4,r1,r2,r3,r4;
int main()
{
    //freopen("1.in","r",stdin);
    //freopen("3.txt","w",stdout);
    long long t;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld %lld %lld %lld %lld %lld %lld %lld",&l1,&r1,&l2,&r2,&l3,&r3,&l4,&r4);
        long long ans = (r1 - l1 + 1) * (r2 - l2 + 1)%mod;
        ans = ans * (r3 - l3 + 1)%mod;
        ans = ans * (r4 - l4 + 1)%mod;
        long long  l,r,temp,rr,ll;
        // 1 == 2
        l = max(l1,l2);r = min(r1,r2);
        if(r >= l)
            ans=((ans-(r-l+1)*(r3-l3+1)%mod*(r4-l4+1)%mod)%mod+mod)%mod;
        //2 == 3
        l = max(l3,l2);r = min(r3,r2);
        if(r >= l)
            ans = ((ans - (r - l + 1) * (r1 - l1 + 1) % mod * (r4 - l4 + 1)%mod)%mod + mod) % mod;
        //3 == 4
        l = max(l3,l4);r = min(r3,r4);
        if(r >= l)
            ans = ((ans - (r - l + 1) * (r1 - l1 + 1) % mod * (r2 - l2 + 1)%mod)%mod + mod) % mod;
        // 4 == 1
        l = max(l1,l4);r = min(r1,r4);
        if(r >= l)
            ans = ((ans - (r - l + 1) * (r3 - l3 + 1) % mod * (r2 - l2 + 1)%mod)%mod + mod) % mod;

        //1 == 2 == 3
        r = min(min(r1,r2),r3),l = max(max(l1,l2),l3);
        if(r >= l)
            ans = (ans + (r - l + 1) * (r4 - l4 + 1)%mod)%mod;
        //2 == 3 == 4
        r = min(min(r4,r2),r3),l = max(max(l4,l2),l3);
        if(r >= l)
            ans = (ans + (r - l + 1) * (r1 - l1 + 1)%mod)%mod;
        //1 == 3 == 4
        r = min(min(r1,r4),r3),l = max(max(l1,l4),l3);
        if(r >= l)
            ans = (ans + (r - l + 1) * (r2 - l2 + 1)%mod)%mod;
        //1 == 2 == 4
        r = min(min(r1,r2),r4),l = max(max(l1,l2),l4);
        if(r >= l)
            ans = (ans + (r - l + 1) * (r3 - l3 + 1)%mod)%mod;
        //1 ==2 3 == 4
        r = min(r1,r2),rr = min(r3,r4),l = max(l1,l2),ll = max(l3,l4);
        if(r >= l && rr >= ll)
            ans = (ans + (r - l + 1) * (rr - ll + 1)%mod)%mod;
        //2 == 3 4 == 1
        r = min(r3,r2),rr = min(r4,r1),l = max(l2,l3),ll = max(l4,l1);
        if(r >= l && rr >= ll)
            ans = (ans + (r - l + 1) * (rr - ll + 1)%mod)%mod;
        l = max(max(max(l1,l2),l3),l4),r = min(min(min(r1,r2),r3),r4);
        if(r >= l)
        ans = ((ans - (3 * (r - l + 1))%mod)%mod + mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}