莫比乌斯反演练习题。
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long ll;
int T, b, d, a, c, k, mu[100005], pri[100005], cnt;
bool isp[100005];
ll ans;
void shai(){
memset(isp, true, sizeof(isp));
isp[0] = isp[1] = false;
mu[1] = 1;
for(int i=2; i<=100000; i++){
if(isp[i]) pri[++cnt] = i, mu[i] = -1;
for(int j=1; j<=cnt; j++){
if(i*pri[j]>100000) break;
isp[i*pri[j]] = false;
if(i%pri[j]==0){
mu[i*pri[j]] = 0;
break;
}
mu[i*pri[j]] = -mu[i];
}
}
}
ll getAns(int l, int r){
ll tmp=0;
for(int i=1; i<=l; i++)
tmp += mu[i] * (ll)(l/i) * (ll)(r/i);
return tmp;
}
int main(){
cin>>T;
shai();
for(int i=1; i<=T; i++){
ans = 0;
scanf("%d %d %d %d %d", &a, &b, &c, &d, &k);
if(k){
b /= k; d /= k;
ans = getAns(min(b,d), max(b,d)) - getAns(min(b,d), min(b,d))/2;
}
printf("Case %d: %lld
", i, ans);
}
return 0;
}