hdu 4135 容斥原理

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又搞了一道容斥原理。

题目：求【1，n】区间对m互质的数有多少个？

#include<iostream>#include<vector>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL __int64const int maxn = 1e5+8;LL a[maxn],cn,numpri[maxn],vis[maxn],dis[maxn];LL n,m;LL f[maxn];void getprim(){    cn = 0;    for(LL i=2;i<10000;i++){        if(!dis[i]){            numpri[cn++] = i;            for(LL j=2*i;j<=10000;j+=i)                dis[j] = 1;        }    }}LL getans(LL Max,LL d){    vector <int > v;    LL sum = 0;        for(LL j=0;j<cn&&numpri[j] * numpri[j] <=d;j++){            if(d % numpri[j] == 0){                v.push_back(numpri[j]);                while(d % numpri[j] == 0)d /= numpri[j];            }        }        if(d!=1)v.push_back(d);        int len = v.size();        //cout << len <<endl;        for(int j=0;j<(1<<len);j++){            LL op =0,ans = 1;            for(int k=0;k<len;k++){                if((1<<k)&j){                    op++;                    ans*=v[k];                }            }            if(op&1)sum-= Max/ans;            else sum += Max/ans;        }    return sum;}int su = 0;int main(){    LL T;    LL a,b,c;    scanf("%I64d",&T);    getprim();    while(T--){        cin >> a>>b>>c;       // cout << getans(b,c)<<endl;       printf("Case #%d: %I64d\n",++su,getans(b,c) - getans(a-1,c));    }}

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