HDU 2620 - Ice Rain (数学)

题意

求 ∑ni=1kmodi

思路

amodb=a−b∗⌊ab⌋

所以ans=k∗n−1∗⌊k1⌋−2∗⌊k2⌋−...−n∗⌊kn⌋

可以成段统计。

计算出某个 ⌊ka⌋=⌊kb⌋的区间a b

代码

#include <stack>#include <cstdio>#include <list>#include <cassert>#include <set>#include <iostream>#include <string>#include <vector>#include <queue>#include <functional>#include <cstring>#include <algorithm>#include <cctype>#include <string>#include <map>#include <cmath>using namespace std;#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define Lowbit(x) ((x) & (-x))#define MP(a, b) make_pair(a, b)#define MS(arr, num) memset(arr, num, sizeof(arr))#define PB push_back#define X first#define Y second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define LC rt << 1, l, mid#define RC rt << 1|1, mid + 1, r#define LRT rt << 1#define RRT rt << 1|1const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const double eps = 1e-8;const int MAXN = 2e3 + 10;const int MOD = 1e9 + 7;const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };const int hash_size = 4e5 + 10;int cases = 0;typedef pair<int, int> pii;int main(){    LL n, k;    while (cin >> n >> k)    {        LL ans = n*k;        if (n > k) n = k;        for (int i = 1; i <= n; )        {            LL a = k / i;            LL ed = k / a;            if (ed > n) ed = n;            ans -= (i + ed) * (ed - i + 1) / 2 * a;            i = ed+1;        }        cout << ans << endl;    }    return 0;}