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| #include <bits/stdc++.h>
#define LL long long
using namespace std;
// 求n的质因数
void FindPrimeFactor(int n, int a[], int &cnt)
{
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
a[cnt++] = i;
while (n % i == 0) n /= i;
}
}
if (n > 1) a[cnt++] = n;
return;
}
// 复杂度也是2^k
LL Solve2(LL n, int a[], int cnt)
{
LL res = 0;
int t = 1, que[1005] = { -1}; // 依据容斥原理，que[0]的值是-1
for (int i = 0; i < cnt; i++) { // 比如2 3 4，形成-1 2 3 -6 4 -8 -12 24
int k = t;
for (int j = 0; j < k; j++) {
que[t++] = que[j] * a[i] * (-1); // 难点
/*乘以-1和-1^k是一样的效果
重点在第一个for循环，新加进去的数和前面形成组合子串*/
}
}
cout << "t = " << t << endl;
for (int i = 1; i < t; i++) {
cout << que[i] << endl;
res += n / que[i];
}
return res;
}
int solve (int n, int r)
{
/*求质数*/
vector<int> p;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
p.push_back (i);
while (n % i == 0) n /= i;
}
}
if (n > 1) p.push_back (n);
int sum = 0; // 非互质的个数
for (int msk = 1; msk < (1 << (int)p.size()); msk++) { // 子串有2^k种可能
int mult = 1, bits = 0;
/*每种可能用bit位表示，最多k位，然后看bit位上是否形成组合*/
for (int i = 0; i < (int)p.size(); ++i) {
if (msk & (1 << i)) {
++bits;
mult *= p[i];
}
}
int cur = r / mult;
//cout << cur << endl;
if (bits % 2 == 1) sum += cur;
else sum -= cur;
}
return r - sum;
}
/*
题目：给定一个区间[A,B],找出在这个区间内与给定的N互质的整数的个数。
T(0 < T <= 100)the number of test cases
each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 10^15) and (1 <=N <= 10^9).
直接求互质复杂度高，可以用素数筛选法求非互质的数。
求[1,B]-[1,A-1]。
*/
// 直接算一定有错误，必须使用容斥原理
LL Solve(LL n, int a[], int cnt)
{
LL res = 0;
for (int i = 0; i < cnt; i++) {
for (int j = 1; ; j++) {
if (j * a[i] > n) {
res += j - 1;
break;
}
}
}
return res;
}
/*
N = 12 --- 2 3
18 2 4 6 8 10 12 14 16 18
3 6 9 12 15 18
会返回结果为15.
正确答案是12.
*/
int main()
{
const int maxn = 1005;
int a[maxn] = {2, 3, 4}, cnt = 3;
//FindPrimeFactor(12, a, cnt);
//cout << Solve2(18, a, cnt) << endl;
cout << solve(12, 18) << endl;
__int64 i, T, x, y, n, sum;
while(scanf("%I64d", &T) != EOF) {
for(i = 1; i <= T; i++) {
scanf("%I64d%I64d%I64d", &x, &y, &n);
//sum = y - haha(y) - (x - 1 - haha(x - 1)); //由于区间是[x,y],求出[1,y]的互质个数，再减去[1,x-1]的互质个数
printf("Case #%I64d: ", i);
printf("%I64d\n", sum);
}
}
return 0;
}
|
v1.5.2