【埃式筛法】My pretty girl Noora CodeForces

Think:
1知识点：埃式筛法+递推
2题意：n个女生参加选美大赛，每次分成n/x组，每组进行x*(x-1)/2次比较，然后晋级n/x名女生，不断分组，不断选拔，最终只留下1名女生作为冠军，f[n]表示n名女生决出冠军的最少比较次数(通过改变x可影响f[x]).
输入t, l , r (1 ≤ t < 1e9 + 7, 2 ≤ l ≤ r ≤ 5·1e6)
求解:t^0·f(l) + t^1·f(l + 1) + … + t^(r - l)·f(r)

3借鉴前辈解题思路：

4反思：
1>方法要立足理解题意，一开始未认真分析就使用快速幂进而超时
2>t^()未注意及时取模，导致Wrong Answer

vjudge题目链接

建议参考博客1
建议参考博客2

以下为Time Limit Exceeded代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int mod = 1000000007;const int N = 5001400;int prim[N], tp, link[N];LL f[N];void get_primes();void get_f();LL find_primes(LL x);LL power(LL x, int k, int mod);int main(){    get_primes();    get_f();    LL x, sum;    int l, r, k, i;    while(~scanf("%lld %d %d", &x, &l, &r)){        sum = 0;        for(k = 0, i = l; i <= r; i++, k++){            sum = (sum + power(x, k, mod)*f[i]) % mod;        }        printf("%lld\n", sum);    }    return 0;}void get_primes(){    tp = 0;    memset(prim, 0, sizeof(prim));    prim[1] = 1, prim[2] = 0;    for(LL i = 2; i < N; i++){        if(!prim[i]){            link[tp++] = i;            for(LL j = i*i; j < N; j += i){                prim[j] = 1;            }        }    }}void get_f(){    for(LL i = 2; i < N; i++){        if(!prim[i])            f[i] = i*(i-1)/2;        else {            LL t = find_primes(i);            f[i] = (i/t)*f[t] + f[i/t];        }        f[i] %= mod;    }}LL find_primes(LL x){    LL ans;    for(LL i = 0; i < tp; i++){        if(x % link[i] == 0){            ans = link[i];            break;        }    }    return ans;}LL power(LL x, int k, int mod){    LL ans = 1;    while(k){        if(k & 1)            ans = (ans*x) % mod;        x = (x*x) % mod;        k >>= 1;    }    return ans;}

以下为Wrong Answer代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int mod = 1000000007;const int N = 5001400;int prim[N], tp, link[N];LL f[N];void get_primes();void get_f();LL find_primes(LL x);int main(){    get_primes();    get_f();    LL x, sum, po;    int l, r, i;    while(~scanf("%lld %d %d", &x, &l, &r)){        sum = 0, po = 1;        for(i = l; i <= r; i++){            sum = (sum + po*f[i]) % mod;            po *= x;        }        printf("%lld\n", sum);    }    return 0;}void get_primes(){    tp = 0;    memset(prim, 0, sizeof(prim));    prim[1] = 1, prim[2] = 0;    for(LL i = 2; i < N; i++){        if(!prim[i]){            link[tp++] = i;            for(LL j = i*i; j < N; j += i){                prim[j] = 1;            }        }    }}void get_f(){    for(LL i = 2; i < N; i++){        if(!prim[i])            f[i] = i*(i-1)/2;        else {            LL t = find_primes(i);            f[i] = (i/t)*f[t] + f[i/t];        }        f[i] %= mod;    }}LL find_primes(LL x){    LL ans;    for(LL i = 0; i < tp; i++){        if(x % link[i] == 0){            ans = link[i];            break;        }    }    return ans;}

以下为Accepted代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int mod = 1000000007;const int N = 5001400;int prim[N], tp, link[N];LL f[N];void get_primes();void get_f();LL find_primes(LL x);int main(){    get_primes();    get_f();    LL x, sum, po;    int l, r, i;    while(~scanf("%lld %d %d", &x, &l, &r)){        sum = 0, po = 1;        for(i = l; i <= r; i++){            sum = (sum + po*f[i]) % mod;            po = (po*x) % mod;        }        printf("%lld\n", sum);    }    return 0;}void get_primes(){    tp = 0;    memset(prim, 0, sizeof(prim));    prim[1] = 1, prim[2] = 0;    for(LL i = 2; i < N; i++){        if(!prim[i]){            link[tp++] = i;            for(LL j = i*i; j < N; j += i){                prim[j] = 1;            }        }    }}void get_f(){    for(LL i = 2; i < N; i++){        if(!prim[i])            f[i] = i*(i-1)/2;        else {            LL t = find_primes(i);            f[i] = (i/t)*f[t] + f[i/t];        }        f[i] %= mod;    }}LL find_primes(LL x){    LL ans;    for(LL i = 0; i < tp; i++){        if(x % link[i] == 0){            ans = link[i];            break;        }    }    return ans;}