Educational Codeforces Round 5 E. Sum of Remainders (思维题)

题目链接：http://codeforces.com/problemset/problem/616/E

题意很简单就不说了。

因为n % x = n - n / x * x

所以答案就等于 n * m - （n/1*1 + n/2*2 ... n/m*m)

在根号n复杂度枚举x，注意一点当m>n时，后面一段加起来就等于0，就不用再枚举了。

中间一段x1 ~ x2 的n/x可能相等，所以相等的一段等差数列求和。

 //#pragma comment(linker, "/STACK:102400000, 102400000")
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 typedef __int64 LL;
 typedef pair <int, int> P;
 const int N = 1e5 + ;
 vector <LL> myset; //存储x
 LL mod = 1e9 + ;

 int main()
 {
     LL n, m;
     scanf("%lld %lld", &n, &m);
     LL k = min(m, n);
     myset.push_back(k);
     for(LL i = ; i*i <= n; ++i) {
         myset.push_back(i);
         if(i * i != n) {
             myset.push_back(n/i);
         }
     }
     sort(myset.begin(), myset.end());
     LL ans = (n%mod)*(m%mod)%mod, cnt = ;
     for(LL i = ; i < myset.size() && myset[i] <= k; ++i) {
         LL temp = myset[i];
         if(cnt) {
             LL num;
             if((temp - cnt) % )
                 num = ((temp + cnt + ) /  % mod) * ((temp - cnt) % mod) % mod;
             else
                 num = ((temp - cnt) /  % mod) * ((temp + cnt + ) % mod) % mod;
             num = ((n / temp) % mod * num) % mod;
             ans = ((ans - num) % mod + mod) % mod;
         }
         else {
             ans = (ans - n) % mod;
         }
         cnt = temp;
     }
     printf("%lld\n", (ans + mod) % mod);
     return ;
 }