Codeforces 272 div2

C

设 m = x%b, x = km*b+m = m(kb+1), 1<=m<b, 所以可以先算出 sum_m = m(m-1)/2

然后枚举k( 1<=k<=a )即可。注意取MOD，不然很容易爆LL。。。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <cassert>
#include <algorithm>
#include <cmath>
#include <set>
#include <list>
#include <map>
#include <limits>
using namespace std;

#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))
#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))

#define MAXN 10000
#define MAXM 10000
#define MOD 1000000007

#define PI 3.1415926535897932384626433832795
#define HALF_PI 1.5707963267948966192313216916398

typedef long long LL;
typedef vector<int> veci;
typedef vector<pair<int, int> > vect;
typedef pair<int, int> pairi;

const double maxdouble = numeric_limits<double>::max();
const double eps = 1e-10;
const int INF = 0x7FFFFFFF;

int main() {
LL a, b;
cin >> a >> b;

LL m = (1+b-1)*(b-1)/2%MOD;
LL sum = 0;
REP(k, 1, a) {
sum = (sum+m*(k*b%MOD+1)%MOD)%MOD;
}
cout << sum;
return 0;
}D
枚举i从0到n-1 然后对于每个i S = {(6i+1)*k, (6i+2)*k, (6i+3)*k, (6i+5)*k}

S里面的4个数全部除以k后全部两两互质，4个质数的间隔最小为6，即1 2 3 5