時間限制: 10 Sec 記憶體限制: 128 MB
題目描述
Given l1,r1,l2,r2,l3,r3,l4,r4, please count the number of four-tuples (x1,x2,x3,x4) such that li≤ xi≤ ri and x1≠x2,x2≠x3,x3≠x4,x4≠x1. The answer should modulo 10^9+7 before output.
輸入
The input consists of several test cases. The first line gives the number of test cases, T(1≤ T≤ 10^6).
For each test case, the input contains one line with 8 integers l1,r1,l2, r2, l3,r3,l4,r4(1≤ li≤ ri≤ 10^9)
輸出
For each test case, output one line containing one integer, representing the answer.
樣例輸入
1
1 1 2 2 3 3 4 4
樣例輸出
1

題意：
給你四個區間，要求每個區間選一個數組成一個四元組（x1,x2,x3,x4$x_1,x_2,x_3,x_4$）,要求
x1≠x2,x2≠x3,x3≠x4,x4≠x1$x_1\ne x_2,x_2\ne x_3,x_3\ne x_4,x_4\ne x_1$

solution
1.先將四個區間長度的乘積作為答案
2.分別減去 x1=x2,x2=x3,x3=x4,x4=x1$x_1=x_2,x_2=x_3,x_3=x_4,x_4=x_1$ 四種情況的組合數量（每種情況中未提及的變數在其區間中任選，即統計答案時直接乘區間長度）
3.因為減去 x1=x2$x_1=x_2$ 和 x2=x3$x_2=x_3$ 時會重複減去 x1=x2=x3$x_1=x_2=x_3$ 的情況，所以要加回來
類似的還有

#define IN_LB() freopen("C:\\Users\\acm2018\\Desktop\\in.txt","r",stdin)
#define OUT_LB() freopen("C:\\Users\\acm2018\\Desktop\\out.txt"
 
,"w",stdout)
#define IN_PC() freopen("C:\\Users\\hz\\Desktop\\in.txt","r",stdin)
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int T;
const ll MOD = 1e9+7;
int main() {
//    IN_LB();
    scanf("%d",&T);
    while(T--) {
        ll l1,r1,l2,r2,l3,r3,l4,r4;
        ll maxl_1,minr_1,maxl_2,minr_2;
        scanf("%lld%lld%lld%lld%lld%lld%lld%lld",&l1,&r1,&l2,&r2,&l3,&r3,&l4,&r4);
        ll ans = (r1-l1+1)*(r2-l2+1)%MOD;
        ans = ans*(r3-l3+1)%MOD;
        ans = ans*(r4-l4+1)%MOD;
        //1==2
        maxl_1 = max(l1,l2);
        minr_1 = min(r1,r2);
        if(maxl_1<=minr_1) {
            ans = ((ans-(minr_1-maxl_1+1)*(r3-l3+1)%MOD*(r4-l4+1)%MOD)%MOD+MOD)%MOD;
        }
        //2==3
        maxl_1 = max(l2,l3);
        minr_1 = min(r2,r3);
        if(maxl_1<=minr_1) {
            ans = ((ans-(minr_1-maxl_1+1)*(r4-l4+1)%MOD*(r1-l1+1)%MOD)%MOD+MOD)%MOD;
        }
        //3==4
        maxl_1 = max(l3,l4);
        minr_1 = min(r3,r4);
        if(maxl_1<=minr_1) {
            ans = ((ans-(minr_1-maxl_1+1)*(r1-l1+1)%MOD*(r2-l2+1)%MOD)%MOD+MOD)%MOD;
        }
        //1==4
        maxl_1 = max(l1,l4);
        minr_1 = min(r1,r4);
        if(maxl_1<=minr_1) {
            ans = ((ans-(minr_1-maxl_1+1)*(r2-l2+1)%MOD*(r3-l3+1)%MOD)%MOD+MOD)%MOD;
        }
        //1==2&&2==3
        maxl_1 = max(l1,max(l2,l3));
        minr_1 = min(r1,min(r2,r3));
        if(maxl_1<=minr_1) {
            ans = (ans+(minr_1-maxl_1+1)*(r4-l4+1)%MOD)%MOD;
        }
        //1==2&&1==4
        maxl_1 = max(l1,max(l2,l4));
        minr_1 = min(r1,min(r2,r4));
        if(maxl_1<=minr_1) {
            ans = (ans+(minr_1-maxl_1+1)*(r3-l3+1)%MOD)%MOD;
        }
        //1==2&&3==4
        maxl_1 = max(l1,l2);
        minr_1 = min(r1,r2);
        maxl_2 = max(l3,l4);
        minr_2 = min(r3,r4);
        if(minr_1>=maxl_1&&minr_2>=maxl_2){
            ans = (ans+(minr_1-maxl_1+1)*(minr_2-maxl_2+1)%MOD)%MOD;
        }
        //2==3&&3==4
        maxl_1 = max(l2,max(l3,l4));
        minr_1 = min(r2,min(r3,r4));
        if(maxl_1<=minr_1) {
            ans = (ans+(minr_1-maxl_1+1)*(r1-l1+1)%MOD)%MOD;
        }
        //2==3&&1==4
        maxl_1 = max(l3,l2);
        minr_1 = min(r3,r2);
        maxl_2 = max(l1,l4);
        minr_2 = min(r1,r4);
        if(minr_1>=maxl_1&&minr_2>=maxl_2){
            ans = (ans+(minr_1-maxl_1+1)*(minr_2-maxl_2+1)%MOD)%MOD;
        }
        //3==4&&1==4
        maxl_1 = max(l1,max(l3,l4));
        minr_1 = min(r1,min(r3,r4));
        if(maxl_1<=minr_1) {
            ans = (ans+(minr_1-maxl_1+1)*(r2-l2+1)%MOD)%MOD;
        }
        //1==2&&2==3&&3==4
        maxl_1 = max(max(l1,l2),max(l3,l4));
        minr_1 = min(min(r1,r2),min(r3,r4));
        if(maxl_1<=minr_1){
            ans = ((ans-(minr_1-maxl_1+1)*3)%MOD+MOD)%MOD;
        }
        printf("%lld\n",ans);
    }
    return 0;
}