标签：mes pen present 内存 splay hat close sts ber
时间限制: 10 Sec 内存限制: 128 MB
题目描述
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.
输入
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)
输出 For each test case, output one line containing one integer, representing the answer.
样例输入
1 1 1 2 2 3 3 4 4
样例输出
1
容斥定理，先把每一部分的交集求出来，然后根据公式求。
AC代码；
include <bits/stdc++.h> using namespace std; int bijiao[20];//每次比较时记录四个坐标 int jiaoji(int x1,int x2,int y1,int y2)//求交集 {if(x2<y1||x2>y2) { return 0; } else { bijiao[0]=x1; bijiao[1]=x2; bijiao[2]=y1; bijiao[3]=y2; sort(bijiao,bijiao+4); return bijiao[2]-bijiao[1]+1; } } int main() { int T; int a[5][2]; long long int b[100][100]; scanf("%d",&T); while(T--) { scanf("%d%d%d%d%d%d%d%d",&a[0][0],&a[0][1],&a[1][0],&a[1][1],&a[2][0],&a[2][1],&a[3][0],&a[3][1]); b[0][1]=a[0][1]-a[0][0]+1; b[0][2]=a[1][1]-a[1][0]+1; b[0][3]=a[2][1]-a[2][0]+1; b[0][4]=a[3][1]-a[3][0]+1; b[1][2]=jiaoji(a[0][0],a[0][1],a[1][0],a[1][1]); b[1][3]=jiaoji(a[0][0],a[0][1],a[2][0],a[2][1]); b[1][4]=jiaoji(a[0][0],a[0][1],a[3][0],a[3][1]); b[2][3]=jiaoji(a[1][0],a[1][1],a[2][0],a[2][1]); b[2][4]=jiaoji(a[1][0],a[1][1],a[3][0],a[3][1]); b[3][4]=jiaoji(a[2][0],a[2][1],a[3][0],a[3][1]); if(b[1][2]==0) { b[12][34]=0; } else { int p=jiaoji(a[0][0],a[0][1],a[1][0],a[1][1]); b[12][34]=jiaoji(bijiao[2],bijiao[1],a[2][0],a[2][1]); } if(b[1][3]==0) { b[13][42]=0; } else { int p=jiaoji(a[0][0],a[0][1],a[2][0],a[2][1]); b[13][42]=jiaoji(bijiao[2],bijiao[1],a[3][0],a[3][1]); } if(b[1][2]==0) { b[12][43]=0; } else { int p=jiaoji(a[0][0],a[0][1],a[1][0],a[1][1]); b[12][43]=jiaoji(bijiao[2],bijiao[1],a[3][0],a[3][1]); } if(b[2][3]==0) { b[23][41]=0; } else { int p=jiaoji(a[1][0],a[1][1],a[2][0],a[2][1]); b[23][41]=jiaoji(bijiao[2],bijiao[1],a[3][0],a[3][1]); } if(b[12][34]==0) { b[11][11]=0; } else { int p=jiaoji(a[0][0],a[0][1],a[1][0],a[1][1]); b[12][34]=jiaoji(bijiao[2],bijiao[1],a[2][0],a[2][1]); b[11][11]=jiaoji(bijiao[2],bijiao[1],a[3][0],a[3][1]); } long long int s; s=(((b[0][4]*b[0][1]%1000000007)*b[0][2]%1000000007)*b[0][3]%1000000007); s=(s-((b[1][2]*b[0][3]%1000000007)*b[0][4]%1000000007)+1000000007)%1000000007; s=(s-((b[1][4]*b[0][2]%1000000007)*b[0][3]%1000000007)-((b[2][3]*b[0][1]%1000000007)*b[0][4]%1000000007)+1000000007)%1000000007; s=(s-((b[3][4]*b[0][1]%1000000007)*b[0][2]%1000000007)+1000000007)%1000000007; s=(s+(b[12][34]*b[0][4])%1000000007)%1000000007; s=(s+b[1][2]*b[3][4]%1000000007)%1000000007; s=(s+b[2][3]*b[1][4]%1000000007)%1000000007; s=(s+b[12][43]*b[0][3])%1000000007; s=(s+b[13][42]*b[0][2])%1000000007; s=(s+b[23][41]*b[0][1])%1000000007; s=(s-b[11][11]*3+1000000007)%1000000007; printf("%lld\n",s); } return 0; }
标签：mes pen present 内存 splay hat close sts ber
原文地址：https://www.cnblogs.com/xiaolaji/p/9135741.html
