Matt loves letter L.
A point set P is (a, b)-L if and only if there exists x, y satisfying:
P = {(x, y), (x + 1, y), . . . , (x + a, y), (x, y + 1), . . . , (x, y + b)}(a, b ≥ 1)
A point set Q is good if and only if Q is an (a, b)-L set and gcd(a, b) = 1.
Matt is given a point set S. Please help him find the number of ordered pairs of sets (A, B) such that:
Input
The first line contains only one integer T , which indicates the number of test cases.
For each test case, the first line contains an integer N (0 ≤ N ≤ 40000), indicating the size of the point set S.
Each of the following N lines contains two integers xi, yi, indicating the i-th point in S (1 ≤ xi, yi ≤ 200). It’s guaranteed that all (xi, yi) would be distinct.
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.
这道题DP计数，十分有意义。
dp[i][j]：表示a∈[1,i],b∈[1,j],sigma(a,b)==1
sum[i][j]：表示竖直部分穿过点(i,j)的L的个数
看代码就能懂了……
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N=210;
typedef long long LL;
LL map[N][N],st[N],S,M;
LL f[N][N],dp[N][N],sum[N][N];
/*f[i][j]:[1,j]中与i互质的数的个数*/
LL dwn[N][N],rht[N][N];int T,cas,k,x,y;
LL Gcd(LL a,LL b){return b?Gcd(b,a%b):a;}
void Prepare(){
for(int i=1;i<=200;i++)
for(int j=1;j<=200;j++){
f[i][j]=f[i][j-1]+(Gcd(i,j)==1);
dp[i][j]=dp[i-1][j]+f[i][j];
}
}
void Init(){S=M=0;
memset(map,0,sizeof(map));
memset(sum,0,sizeof(sum));
memset(dwn,0,sizeof(dwn));
memset(rht,0,sizeof(rht));
}
int main(){
Prepare();
scanf("%d",&T);
while(T--){
Init();scanf("%d",&k);
while(k--){scanf("%d%d",&x,&y);map[x][y]=1;}
for(int i=200;i>=1;i--)
for(int j=200;j>=1;j--){
if(!map[i][j])continue;
if(map[i+1][j])dwn[i][j]=dwn[i+1][j]+1;
if(map[i][j+1])rht[i][j]=rht[i][j+1]+1;
}
for(int i=1;i<=200;i++)
for(int j=1;j<=200;j++){
if(!map[i][j])continue;
memset(st,0,sizeof(st));
for(int k=1;k<=dwn[i][j];k++)
st[k]=f[k][rht[i][j]];
for(int k=dwn[i][j];k>=1;k--)
st[k-1]+=st[k];
for(int k=0;k<=dwn[i][j];k++)
sum[i+k][j]+=st[k];
S+=st[0];
}
for(int i=1;i<=200;i++)
for(int j=1;j<=200;j++){
if(!dwn[i][j])continue;
if(!rht[i][j])continue;
LL tot=dp[dwn[i][j]][rht[i][j]];
LL calc=sum[i][j]-tot;
for(int k=1;k<=rht[i][j];k++){
calc+=sum[i][j+k];
M+=2*calc*f[k][dwn[i][j]];
}
M+=tot*tot;
}
printf("Case #%d: %lld\n",++cas,S*S-M);
}
return 0;
}