HDU 5528 Count a × b

Count a * b Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1571    Accepted Submission(s): 523   Problem Description Marry likes to count the number of ways to choose two non-negative integers a and b less than m to make a×b mod m≠0 . Let's denote f(m) as the number of ways to choose two non-negative integers a and b less than m to make a×b mod m≠0 . She has calculated a lot of f(m) for different m , and now she is interested in another function g(n)=∑m|nf(m) . For example, g(6)=f(1)+f(2)+f(3)+f(6)=0+1+4+21=26 . She needs you to double check the answer.   Give you n . Your task is to find g(n) modulo 264 .     Input The first line contains an integer T indicating the total number of test cases. Each test case is a line with a positive integer n . 1≤T≤20000 1≤n≤109     Output For each test case, print one integer s , representing g(n) modulo 264 .     Sample Input   2 6 514     Sample Output   26 328194     这道题就是求     推到这一步,然后就有了一种sqrt(n)的解法 然而T=20000 于是成功卡常 参考唯一分解定理   于是现在可以先筛一个素数表 复杂度sqrt(n)以内的质数  sqrt(n)/ln(n) 比上一种方法快几倍 这道题时间卡得很紧   然而，成功爆unsigned long long    在这里， 引入__int128成功卡过   网上还有其他的防止溢出的方法 #include using namespace std; typedef unsigned long long ull; const int MAX=100010; long long su[MAX],cnt; bool isprime[MAX]; void prime() { cnt=1; memset(isprime,1,sizeof(isprime)); isprime[0]=isprime[1]=0; for(long long i=2; i<=MAX; i++) { if(isprime[i]) su[cnt++]=i; for(long long j=1; jn) break; int k=0; ull tp=su[i]; while(n%su[i]==0) { k++; tp*=su[i]; n/=su[i]; } //tp=tp*tp-1; if(k) { io=(__int128)tp*tp-1;//爆ull ans1=ans1*(ull)(io/((ull)su[i]*su[i]-1)); ans2=ans2*(k+1); } } if(n!=1) { ans1=ans1*((ull)n*n+1); ans2=ans2*2; } printf("%llu\n",ans1-ans2); } return 0; }