18年省赛 F Four-tuples 容斥原理

 

 

F-Four-tuples_“浪潮杯”第九届山东省ACM大学生程序设计竞赛重现赛(重现赛)（重现赛）@aquamoon_zjx (nowcoder.com)

 

 

写公式的时候, 可以看着图中的某一块, 数它加了多少,减了多少来保证正确性

 

 

 代码

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
const LL mod = 1e9+7;
int l[5], r[5];

LL f2(int a,int b)
{
    LL lt=max(l[a],l[b]);
    LL rt=min(r[a],r[b]);
    return rt-lt+1>0?rt-lt+1:0;
}
LL f3(int a,int b,int c)
{
    LL lt=max(l[a],max(l[b],l[c]));
    LL rt=min(r[a],min(r[b],r[c]));
    return rt-lt+1>0?rt-lt+1:0;
}
LL f4(int a,int b,int c,int d)
{
    LL lt=max(max(l[a],l[b]),max(l[c],l[d]));
    LL rt=min(min(r[a],r[b]),min(r[c],r[d]));
    return rt-lt+1>0?rt-lt+1:0;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        LL res = 1;
        for(int i = 1; i <=4; i ++)
            scanf("%d %d", &l[i], &r[i]);//用scanf, 否则超时
        
        for(int i = 1; i <=4; i ++)
            res = (res*(r[i]-l[i]+1)%mod)%mod;
            
        res = (res - f2(1,2)%mod*(r[3]-l[3]+1)%mod * (r[4]-l[4]+1)%mod+mod)%mod;
        res = (res - f2(2,3)%mod*(r[1]-l[1]+1)%mod * (r[4]-l[4]+1)%mod+mod)%mod;
        res = (res - f2(3,4)%mod*(r[1]-l[1]+1)%mod * (r[2]-l[2]+1)%mod+mod)%mod;
        res = (res - f2(4,1)%mod*(r[3]-l[3]+1)%mod * (r[2]-l[2]+1)%mod+mod)%mod;
            
        //S1US2时,包括S3US4;  S3US4时,包括S1US2. 这期间重复减去了S1US2且S3US4;  
        res = (res + f2(1,2)%mod*f2(3,4)%mod+mod)%mod;
        res = (res + f2(3,2)%mod*f2(1,4)%mod+mod)%mod;
        
        res = (res + f3(1,2,3)%mod * (r[4]-l[4]+1)%mod)%mod;
        res = (res + f3(2,3,4)%mod * (r[1]-l[1]+1)%mod)%mod;
        res = (res + f3(3,4,1)%mod * (r[2]-l[2]+1)%mod)%mod;
        res = (res + f3(4,1,2)%mod * (r[3]-l[3]+1)%mod)%mod;
        
        res = res - f4(1,2,3,4)%mod*3%mod;//这一块的情况 4-4+4, 最后一共加了4次, 需要再减去3次
        res = (res+mod)%mod;
          
        printf("%lld\n", res);
    }
    return 0;
}

/*
S1+S2+S3+S4 
- S1∩S2-S2∩S3-S3∩S4-S4∩S1 
+ S1∩S2∩S3+S2∩S3∩S4+S3∩S4∩S1+S4∩S1∩S2 
- S1∩S2∩S3∩S4
*/