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[题目链接](http://acm.hdu.edu.cn/showproblem.php?pid=4135)
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https://blog.csdn.net/qq_40507857/article/details/82788443
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> 题意：求区间`[a,b]`中与`m`互素的数字个数
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### 一、题目分析
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考虑`[1,a-1]`和`[1,b]`两个区间与`m`互素的个数，答案就是二者之差(**类似前缀和？**)。
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<font color='red'><b>容斥原理经典问题</b></font>：求`1-n`中与`m`互素的数的个数
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互素的个数等于总数减去不互素的个数,如果$1-n$中某个数与$m$不互素，那么一定可以被$m$的某个因子整除，所以先枚举$m$的所有素因子。
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举个例子，$m=12,n=8$，$12$的素因子是$2，3$，
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$1\sim 8$中有几个是$2$的倍数呢？$S_2=8/2=4$;
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$1\sim 8$中有几个是$3$的倍数呢？$S_3=8/3=2$;
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$1\sim 8$之间与$12$互素的个数是$2+4=6$个数字吗？
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$1\sim 8$之间与$12$不互素的是$2，3，4，6，8$，共$5$个数字，这是因为同为$2$和$3$的倍数的$6$被计算了两次，所以要再减去一次$S_6=8/6=1$,结果是 $$\large 8/2+8/3-8/(2*3)=4+2-1=5$$
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这是经典的容斥原理啊，如果是奇数个组合，那么符号是`+`；如果是偶数个组合，那么符号是`-`；
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### 二、实现代码
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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//返回1-m中与n互素的数的个数
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vector<LL> p;
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LL cal(LL n, LL m) {
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p.clear();
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for (int i = 2; i * i <= n; i++) {
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if (n % i == 0) {
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p.push_back(i);
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while (n % i == 0) n /= i;
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}
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}
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if (n > 1) p.push_back(n); //求n的素因子
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int num = p.size(); //素因子的个数
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LL s = 0; // 1到m中与n不互素的数的个数
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//枚举子集，不能有空集，所以从1开始
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for (LL i = 1; i < 1 << num; i++) { //从1枚举到(2^素因子个数)
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LL cnt = 0;
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LL t = 1;
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for (LL j = 0; j < num; j++) { //枚举每个素因子
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if (i & (1 << j)) { //有第i个因子
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cnt++; //计数
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t *= p[j]; //乘上这个质因子
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}
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}
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//容斥原理
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if (cnt & 1) //选取个数为奇数，加
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s += m / t;
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else //选取个数为偶数，减
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s -= m / t;
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}
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return m - s; //返回1-m中与n互素的数的个数
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}
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int main() {
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//加快读入
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ios::sync_with_stdio(false);
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cin.tie(0);
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cout.tie(0);
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int T, ca = 0;
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cin >> T;
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while (T--) {
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LL m, a, b;
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cin >> a >> b >> m; //求区间[a,b]中与m互素的数字个数
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//计算[1,a-1]之间与m互素的个数
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//计算[1, b]之间与m互素的个数
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LL ans = cal(m, b) - cal(m, a - 1);
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printf("Case #%d: %lld\n", ++ca, ans);
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}
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return 0;
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}
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```
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