0; for(int i=0;i
wh) { printf(0 ); return 0; } pneed = (temp + wh) / 2; mneed = wh - pneed; int ans = 1; for(int i = 0; i < pneed; i++) { ans *= wh; --wh; } for(int i=1; i<= pneed; i++) ans /= i; printf(%.12f ,double(ans)/double(all)); }
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C. Dreamoon and Sums time limit per test 1.5 seconds memory limit per test 256 megabytes
Dreamoon loves summing up something for no reason. One day he obtains two integers a and b occasionally. He wants to calculate the sum of all nice integers. Positive integer x is called nice if 
and 
, where k is some integer number in range [1,?a].
By 
we denote the quotient of integer division of x and y. By 
we denote the remainder of integer division of x and y. You can read more about these operations here: http://goo.gl/AcsXhT.
The answer may be large, so please print its remainder modulo 1?000?000?007 (109?+?7). Can you compute it faster than Dreamoon?
Input
The single line of the input contains two integers a, b (1?≤?a,?b?≤?107).
Output
Print a single integer representing the answer modulo 1?000?000?007 (109?+?7).
Sample test(s) Input
1 1
Output
0
Input
2 2
Output
8
Note
For the first sample, there are no nice integers because 
is always zero.
For the second sample, the set of nice integers is {3,?5}.
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#include
#define ll long long const int mod=1e9 + 7; int main() { ll a,b; scanf(%I64d %I64d, &a, &b); if(b == 1) printf(0 ); else { ll res = (b * (b - 1) / 2) % mod; ll sum = (a * (a + 1) / 2) % mod; ll ans = (a % mod * res % mod) % mod; ll temp = (sum % mod * res % mod) % mod; printf(%I64d , (ans % mod + (temp % mod * b % mod)) % mod); } }
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D. Dreamoon and Sets time limit per test 1 second memory limit per test 256 megabytes
Dreamoon likes to play with sets, integers and 
. 
is defined as the largest positive integer that divides both a and b.
Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sj from S, 
.
Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.
Input
The single line of the input contains two space separated integers n, k (1?≤?n?≤?10?000,?1?≤?k?≤?100).
Output
On the first line print a single integer — the minimal possible m.
On each of the next n lines print four space separated integers representing the i-th set.
Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.
Sample test(s) Input
1 1
Output
5
1 2 3 5
Input
2 2
Output
22
2 4 6 22
14 18 10 16
Note
For the first example it's easy to see that set {1,?2,?3,?4} isn't a valid set of rank 1 since 
.
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#include
int main() { int n, k; scanf(%d %d, &n, &k); int a = 1, b = 2, c = 3, d = 5; printf(%d , (d * k + 6 * k * (n- 1))); a *= k; b *= k; c *= k; d *= k; for(int i = 0; i < n; i++) { printf(%d %d %d %d ,a, b, c, d); a += 6 * k; b += 6 * k; c += 6 * k; d += 6 * k; } }
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