hdu 1695 欧拉函数 + 容斥原理

分析：

本质是求区间内与某个数互质的数的个数，如果区间比这个数小用欧拉函数。比这个数大用全集减去不互质的个数。不互质的个数的计算方法是对当前数进行因数分解，然后求能被p1∪p2∪......∪pn整除的数的个数，显然用容斥。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define pr(x) cout << #x << ": " << x << "  "
#define pl(x) cout << #x << ": " << x << endl;
typedef long long ll;
const int maxn = int(1e5) + 13;
struct jibancanyang
{
int b, d, k;
ll elur[maxn], factors[maxn][8];

void getElur() {
memset(elur, 0, sizeof(elur));
memset(factors, 0, sizeof(factors));
elur[1] = 1;
for (int i = 2; i < maxn; ++i) {
if (!elur[i]) {
for (int j = i; j < maxn; j += i) {
if (!elur[j]) elur[j] = j;
factors[j][++factors[j][0]] = i;
elur[j] = elur[j] * (i - 1) / i;
}
}
// elur[i] += elur[i - 1];
}
}

int doit(int limit, int x) {
ll ret = 0;
for (int s = 1; s < 1 << factors[x][0]; ++s) {
int cur = 1, cnt = 0;
for (int  i = 0; i < factors[x][0]; ++i) {
if (s >> i & 1) ++cnt, cur *= factors[x][i + 1];
}
if (cnt & 1) ret += limit / cur;
else ret -= limit / cur;
}
return (int)ret;
}

void fun() {
int T, a, c, k;
scanf("%d", &T);
for (int cas = 1; cas <= T; ++cas) {
scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
if (k == 0) {
printf("Case %d: 0\n", cas);
continue;
}
b /= k, d /= k;
ll ans = 0;
if (b > d) swap(b, d);
for (int i = d; i >= 1; --i) {
if (b >= i) ans += elur[i];
else {
ans += b - doit(b, i);
}
}
printf("Case %d: %lld\n", cas, ans);
}
}
}ac;

int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif
ac.getElur();
ac.fun();
return 0;
}