[2016-04-21]LightOJ][1341][Aladdin and the Flying Carpet]

  • 时间:2016-04-21 16:45:58 星期四

  • 题目编号:[2016-04-21]LightOJ][1341][Aladdin and the Flying Carpet]

  • 题目大意:

    • 给定数a,b,问满足c×d==a且b⩽c,b⩽d 二元组对数,
    • (c,d)和(d,c)是同种情况
    • c != d

  • 分析:

    • 分解a(唯一分解定理),n=p1a1+p2a2+…+piai+pnan
    • 求约数的个数 cnt=(a1+1)(a2+1)…(ai+1)…(an+1)
    • 因为不考虑c == d的情况,所以直接总的对数直接 n /= 2即可
  1. #include<cstdio>
  2. #include<cstring>
  3. using namespace std;
  4. const int maxn = 1E6 + 10;
  5. int pri[maxn + 1];
  6. void getPri(){
  7.         memset(pri,0,sizeof(pri));
  8.         for(int i = 2 ; i <= maxn ; ++i){
  9.                 if(!pri[i])        pri[++pri[0]] = i;
  10.                 for(int j = 1 ; j <= pri[0] && pri[j] <= maxn/i;++j){
  11.                         pri[pri[j]*i] = 1;
  12.                         if(i % pri[j] == 0)        break;
  13.                 }
  15.         }
  16. }
  17. long long fac[maxn][2];
  18. int fatcnt;
  19. int getFac(long long x){
  20.         fatcnt = 0;
  21.         long long tmp = x;
  22.         for(int i = 1 ; pri[i] <= tmp/pri[i];++i){
  23.                 fac[fatcnt][1] = 0;
  24.                 if(tmp % pri[i] == 0){
  25.                         fac[fatcnt][0] = pri[i];
  26.                         while(tmp % pri[i] == 0){
  27.                                 ++fac[fatcnt][1];
  28.                                 tmp /= pri[i];
  29.                         }
  30.                         ++fatcnt;
  31.                 }
  32.         }
  33.         if(tmp != 1){
  34.                 fac[fatcnt][0] = tmp;
  35.                 fac[fatcnt++][1] = 1;
  36.         }
  37.         return fatcnt;
  38. }
  39. int main(){
  40.         int t,cntcase = 0;
  41.         scanf("%d",&t);
  42.         getPri();
  43.         while(t--){
  44.                 long long a,b;
  45.                 scanf("%lld%lld",&a,&b);
  46.                 if(a < b*b){
  47.                         printf("Case %d: %d\n",++cntcase,0);
  48.                         continue;
  49.                 }
  50.                 getFac(a);
  51.                 int cnt = 1;
  52.                 for(int i = 0; i < fatcnt ; ++i){
  53.                         cnt *= (fac[i][1]+1);
  54.                 }
  55.                 cnt /= 2;
  56.                 for(int i = 1; i < b ; ++i){
  57.                         if(a % i == 0)        --cnt;
  58.                 }
  59.                 printf("Case %d: %d\n",++cntcase,cnt);
  60.         }
  61.         return 0;
  62. }


来自为知笔记(Wiz)


posted on 2016-04-21 17:01  红洋  阅读(140)  评论(0)    收藏  举报

导航