BZOJ2956 模积和

数学题我都不会2333

再次传送门：Orz hzwer

记住[m / i]的分块性质、、、inext = m / (m / i)

 

 1 /**************************************************************
 2     Problem: 2956
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:144 ms
 7     Memory:808 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <algorithm>
12  
13 using namespace std;
14 typedef long long ll;
15 const ll mod = 19940417;
16 const ll p_6 = 3323403;
17  
18 ll n, m, ans;
19  
20 ll sum(ll x, ll y) {
21   return (y - x + 1) * (x + y) / 2 % mod;
22 }
23  
24 ll sum2(ll x) {
25   return x * (x + 1) % mod * (x * 2 + 1) % mod * p_6 % mod;
26 }
27  
28 ll calc(ll x) {
29   ll tmp = 0, i, j;
30   for (i = 1, j; i <= x; i = j + 1) {
31     j = x / (x / i);
32     tmp = (tmp + x * (j - i + 1) % mod - sum(i, j) * (x / i)) % mod;
33   }
34   return (tmp + mod) % mod;
35 }
36  
37 int main() {
38   ll i, j, s1, s2, s3;
39   scanf("%lld%lld", &n, &m);
40   ans = calc(n) * calc(m) % mod;
41   if (n > m) swap(n, m);
42   for (i = 1; i <= n; i = j + 1) {
43     j = min(n / (n / i), m / (m / i));
44     s1 = n * m % mod * (j - i + 1) % mod;
45     s2 = (n / i) * (m / i) % mod * (sum2(j) - sum2(i - 1) + mod) % mod;
46     s3 = (n / i * m + m / i * n) % mod * sum(i, j) % mod;
47     ans = (ans - (s1 + s2 - s3) % mod + mod) % mod;
48   }
49   printf("%lld\n", ans);
50   return 0;
51 }
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posted on 2015-01-30 20:42  Xs酱~  阅读(228)  评论(0)  编辑  收藏  举报