计数问题确实是短板qaq
这个可以先求出符合要求的L，可以通过枚举顶点来求，然后就是求n,m里面互质数的对数，这个其实可以暴力预处理(一开始还想着用反演)。。
然后把总数平方后再减去交叉部分就是答案了。。。
交叉的情况其实很少，就3种
一种是同一个顶点，这个直接减去该顶点的L的平方即可
一种是L的边重叠，这个可以枚举重叠中的顶点，然后一个是在该顶点能构成的L，一个是经过该顶点的L，直接预处理出经过该顶点的L之后乘起来减掉即可。。
一种是2个L的边交叉，只相交于一点，这个可以枚举交叉点，然后预处理横向经过该点的L和纵向经过该点的L，乘起来减掉就行了。。(顺便处理了情况2)
貌似不是很难。。就是思路放不开。。得多接触。。尤其是转化为去算交叉部分那里。。
/**
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ ＞ ＜ ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<stdlib.h>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 205
#define nm 1100005
#define pi 3.1415926535897931
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar() ;
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}
int n,m,_x,_y,ca;
ll ans,d[NM][NM],f[NM][NM],g[NM][NM],b[NM][NM];
bool v[NM][NM];
int main(){
m=200;
inc(i,1,m)inc(j,1,m)b[i][j]=b[i][j-1]+(__gcd(i,j)==1);
int _=read();while(_--){
ans=0;mem(d);mem(f);mem(g);mem(v);
n=read();
inc(i,1,n){_x=read();_y=read();v[_x][_y]++;}
inc(i,1,m)inc(j,1,m)if(v[i][j]){
int x=i,y=j;
inc(k,i+1,m)if(v[k][j])x=k;else break;
inc(k,j+1,m)if(v[i][k])y=k;else break;
inc(k,1,x-i)d[i][j]+=b[k][y-j];
ans+=d[i][j];
ll t=0;
dec(k,y,j+1)t+=b[k-j][x-i],f[i][k]+=t;
t=0;
dec(k,x,i+1)t+=b[k-i][y-j],g[k][j]+=t;
}
ans=sqr(ans);
inc(i,1,m)inc(j,1,m){
ans-=f[i][j]*g[i][j]*2;
ans-=sqr(d[i][j]);
ans-=2*d[i][j]*(f[i][j]+g[i][j]);
}
printf("Case #%d: %lld\n",++ca,ans);
}
return 0;
}
Everlasting LTime Limit: 1000/1000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 420 Accepted Submission(s): 170 Problem Description Matt loves letter L. Input The first line contains only one integer T , which indicates the number of test cases. Output For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the number of pairs. Sample Input 2 6 1 1 1 2 2 1 3 3 3 4 4 3 9 1 1 1 2 1 3 2 1 2 2 2 3 3 1 3 2 3 3 Sample Output Case #1: 2 Case #2: 6 Hint n the second sample, the ordered pairs of sets Matt can choose are: A = {(1, 1), (1, 2), (1, 3), (2, 1)} and B = {(2, 2), (2, 3), (3, 2)} A = {(2, 2), (2, 3), (3, 2)} and B = {(1, 1), (1, 2), (1, 3), (2, 1)} A = {(1, 1), (1, 2), (2, 1), (3, 1)} and B = {(2, 2), (2, 3), (3, 2)} A = {(2, 2), (2, 3), (3, 2)} and B = {(1, 1), (1, 2), (2, 1), (3, 1)} A = {(1, 1), (1, 2), (2, 1)} and B = {(2, 2), (2, 3), (3, 2)} A = {(2, 2), (2, 3), (3, 2)} and B = {(1, 1), (1, 2), (2, 1)} Hence, the answer is 6.Source 2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交） Recommend liuyiding |
