BZOJ1257 洛谷2261 [CQOI2007]余数求和

• Description
  给定n,x，求
  

  ∑i=1nx mod i

• Solution
  

  ==∑i=1nx mod i∑i=1nx−⌊xi⌋×inx−∑i=1n⌊xi⌋×i
  
  后面那一堆显然有大段大段都是一样的，最多不同的个数是n的约数个数。时间复杂度O(n‾√)。

#include <bits/stdc++.h>using namespace std ;#define LL long longLL ans, n, x ;int main() {#ifndef ONLINE_JUDGE    freopen ( "BZOJ1257.in", "r", stdin ) ;    freopen ( "BZOJ1257.out", "w", stdout ) ;#endif    LL i, j, k, lim, l, r ;    scanf ( "%lld%lld", &n, &x ) ;    ans = n*x ;    lim = sqrt(x) ;    for ( i = 1 ; i <= n ; ) {        k = x/i ;        if (!k) break ;        l = i ;        if ( (x+1)%k == 0 ) r = (x+1)/k-1 ;        else r = floor( (x+1)*1.0/k ) ;        r = min(r, n) ;        ans -= (l+r)*(r-l+1)/2*k ;        i = r+1 ;    }    printf ( "%lld\n", ans ) ;      return 0 ;}