BZOJ 1257 [CQOI2007]余数之和sum
给出正整数n和k，计算j(n, k)=k mod 1 + k mod 2 + k mod 3 + … + k mod n的值，其中k mod i表示k除以i的余数。例如j(5, 3)=3 mod 1 + 3 mod 2 + 3 mod 3 + 3 mod 4 + 3 mod 5=0+1+0+3+3=7
(sum_{i=1}^{n} k mod i = sum_{i=1}^{n} k - lfloor frac ki floor i)
(lfloor frac ki floor)的取值很少，在(sqrt k)级别。
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define space putchar(' ')
#define enter putchar('
')
using namespace std;
typedef long long ll;
template <class T>
void read(T &x){
char c;
bool op = 0;
while(c = getchar(), c > '9' || c < '0')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}
ll n, k, ans;
int main(){
read(n), read(k), ans = n * k;
for(ll l = 1, t, r; l <= min(n, k); l = r + 1){
t = k / l, r = min(n, k / t);
ans -= t * (l + r) * (r - l + 1) / 2;
}
write(ans), enter;
return 0;
}