LightOJ1341 Aladdin and the Flying Carpet 约数的个数

1. 利用素因子分解可以求出来共有多少个约数。
2. 给的数据不可能双向卡，所以在此题中要么卡枚举1~b，要么卡b~a，从出题的角度选择了后者，所以要利用素因子分解先求出所有的1~a，再减去1~b，当然，最佳的办法是通过素因子分解的结果去枚举小于b的值，但操作更加复杂。

题目链接：http://acm.hust.edu.cn/vjudge/problem/27049
题意：给定a，b，1<=b<=a<=10^12，求b~a之间的a的约数对的个数

#include<cstdio>#include<iostream>#include<sstream>#include<cstdlib>#include<cmath>#include<cctype>#include<string>#include<cstring>#include<algorithm>#include<stack>#include<queue>#include<set>#include<map>#include<ctime>#include<vector>#include<fstream>#include<list>using namespace std;#define ms(s) memset(s,0,sizeof(s))typedef unsigned long long ULL;typedef long long LL;const int INF = 0x3fffffff;const int N = 1000000;bool primeTable[N+5];vector<int> p;void make_primeTable(){    fill(primeTable,primeTable+N,1);    primeTable[0] = false;    primeTable[1] = false;    int maxed = sqrt(N);    for(int i = 2; i <= maxed; ++i){        if(primeTable[i] == true){            p.push_back(i);            for(int j = i*i; j <= N; j += i)                primeTable[j] = false;        }    }    for(int i = maxed+1; i <= N; ++i)        if(primeTable[i] == true)  p.push_back(i);}LL add_factor(LL n){    if(n == 1)  return 0;    LL s = 1;    for(int i = 0, j = p.size(); i < j; ++i){        LL k = 0;        while(n%p[i] == 0){            k++;            n /= p[i];        }        s *= (k+1);        if(n == 1)  break;    }    if(n > 1)  s *= 2;    s /= 2;    return s;}int main(){//    freopen("F:\\input.txt","r",stdin);//    freopen("F:\\output.txt","w",stdout);//    ios::sync_with_stdio(false);    make_primeTable();    int t;    long long a,b,m;    long long ans;    scanf("%d",&t);    for(int k = 1; k <= t; ++k){        ans = 0;        scanf("%lld%lld",&a,&b);        m = sqrt(a);        if(b > m)            printf("Case %d: 0\n",k);        else{            ans = add_factor(a);            for(int i = 1; i < b; ++i){                if(a%i == 0)                    ans--;            }            printf("Case %d: %lld\n",k,ans);        }    }    return 0;}