2956: 模积和

2019-04-13 14:25发布

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2956: 模积和

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 1284  Solved: 580
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Description

　求∑∑((n mod i)*(m mod j))其中1<=i<=n,1<=j<=m,i≠j。
 　　

Input

第一行两个数n,m。

Output

　　一个整数表示答案mod 19940417的值

Sample Input


3 4

Sample Output

1

样例说明
　　答案为(3 mod 1)*(4 mod 2)+(3 mod 1) * (4 mod 3)+(3 mod 1) * (4 mod 4) + (3 mod 2) * (4 mod 1) + (3 mod 2) * (4 mod 3) + (3 mod 2) * (4 mod 4) + (3 mod 3) * (4 mod 1) + (3 mod 3) * (4 mod 2) + (3 mod 3) * (4 mod 4) = 1

数据规模和约定
　　对于100%的数据n,m<=10^9。

HINT

Source

中国国家队清华集训 2012-2013 第一天
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#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; typedef long long LL; const LL mo = 19940417; int n,m,Inv2,Inv6; int Mul(const LL &x,const LL &y) {return x * y % mo;} int Add(const int &x,const int &y) {return (x + y) % mo;} int Dec(const int &x,const int &y) {return (x - y + mo) % mo;} int sum(int N) { int G = 1 + N; if (G & 1) N >>= 1; else G >>= 1; return Mul(N,G); } int sum2(int N) { int F = N + 1,G = 2 * N + 1; bool f2,f3; f2 = f3 = 0; if (!(N & 1) && !f2) f2 = 1,N >>= 1; if (!(F & 1) && !f2) f2 = 1,F >>= 1; if (!(G & 1) && !f2) f2 = 1,G >>= 1; if (N % 3 == 0 && !f3) f3 = 1,N /= 3; if (F % 3 == 0 && !f3) f3 = 1,F /= 3; if (G % 3 == 0 && !f3) f3 = 1,G /= 3; return Mul(N,Mul(F,G)); } int Work(int N) { int ret = 0; for (int i = 1,last; i <= N; i = last + 1) { int x = N / i; last = N / x; ret = Add(ret,Mul(last - i + 1,N)); ret = Dec(ret,Mul(x,Dec(sum(last),sum(i-1)))); } return ret; } int main() { #ifdef DMC freopen("DMC.txt","r",stdin); #endif cin >> n >> m; int Ans = Mul(Work(n),Work(m)); for (int i = 1,last; i <= min(n,m); i = last + 1) { int x = n / i,y = m / i,tot; last = min(n / x,m / y); tot = Mul(n,Mul(m,last - i + 1)); tot = Dec(tot,Mul(n,Mul(y,Dec(sum(last),sum(i-1))))); tot = Dec(tot,Mul(m,Mul(x,Dec(sum(last),sum(i-1))))); tot = Add(tot,Mul(x,Mul(y,Dec(sum2(last),sum2(i-1))))); Ans = Dec(Ans,tot); } cout << Ans << endl; return 0; }

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