Codeforces 476C - Dreamoon and Sums (数学)
题意
给出a, b，算出有几个数，符合
- 商和余数相除等于k，k属于[1, a]。
思路
引用题解的思路。
If we fix the value of k, and let d = div(x, b), m = mod(x, b), we have :
d = mk
x = db + m
So we have x = mkb + m = (kb + 1) * m.
And we know m would be in range [0, b - 1] because it’s a remainder, so the sum of x of that fixed k would be
Next we should notice that if an integer x is nice it can only be nice for a single particular k because a given x uniquely defines div(x, b) and mod(x, b).
Thus the final answer would be sum up for all individual k: which can be calculated in O(a) and will pass the time limit of 1.5 seconds.
Also the formula above can be expanded to .
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL 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 F first
#define S 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|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 2000 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
int main()
{
LL a, b;
cin >> a >> b;
LL tmp1 = ((1 + a) * a) >> 1;
tmp1 %= MOD;
tmp1 = tmp1 * b % MOD;
tmp1 += a;
LL tmp2 = (b * (b - 1)) >> 1;
tmp2 %= MOD;
LL ans = tmp1 * tmp2 % MOD;
cout << ans << endl;
}