HDU

题意：给定区间[a,b]和n，求区间中与n互素的数的个数, 1≤a,b≤1015,1≤109。

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思路：利用容斥定理求得先求得区间[1,b]与n互素的数的个数，设count(i)表示区间[1,i]中与n互素的数的个数, 那么区间[a,b]中与n互素的数的个数等于count(b)−count(a−1)。详细分析见 求指定区间内与n互素的数的个数 容斥原理

AC代码

#include <cstdio>#include <cmath>#include <cctype>#include <bitset>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 1e4 + 5;LL solve(LL r, int n) {    vector<int>p;    p.clear();    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);    LL sum = 0;    for(int msk = 1; msk < (1<<(p.size())); ++msk) {        LL mult = 1, bits = 0;        for(int i = 0; i < p.size(); ++i) {            if(msk & (1 << i)) {                bits++;                mult *= p[i];            }        }        LL cur = r / mult;        if(bits & 1) sum += cur;        else sum -= cur;    }    return r - sum;}int main() {    LL a, b;    int n;    int T;    scanf("%d", &T);    int kase = 1;    while(T--) {        scanf("%lld%lld%d", &a, &b, &n);        LL x = solve(b, n), y = solve(a-1, n);        //printf("%lld %lld\n", x, y);        printf("Case #%d: %lld\n", kase++, x-y);    }    return 0;} 

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