链接 : http://codeforces.com/contest/476
立即学习“前端免费学习笔记（深入）”；
D题yy，ABC水
立即学习“前端免费学习笔记（深入）”；
A. Dreamoon and Stairs
time limit per test
1 second
memory limit per test
256 megabytes
Dreamoon wants to climb up a stair of n steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer m.
What is the minimal number of steps making him climb to the top of the stairs that satisfies his condition?
Input
The single line contains two space separated integers n, m (0?
Output
Print a single integer ? the minimal number of moves being a multiple of m. If there is no way he can climb satisfying condition print ?-?1 instead.
Sample test(s)
Input
10 2
Output
Input
3 5
Output
-1
Note
For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5.
#include int main(){ int n, m; scanf("%d %d", &n, &m); int temp = n / 2; if(n % 2) temp++; if(temp % m == 0) printf("%d\n", temp); else { temp = temp + m - temp % m; if(temp > n) printf("-1\n"); else printf("%d\n", temp); }} 立即学习“前端免费学习笔记（深入）”；
B. Dreamoon and WiFi
time limit per test
1 second
memory limit per test
256 megabytes
Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them.
Each command is one of the following two types:
- Go 1 unit towards the positive direction, denoted as '+'
- Go 1 unit towards the negative direction, denoted as '-'
But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5).
You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands?
Input
The first line contains a string s1 ? the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}.
The second line contains a string s2 ? the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command.
Lengths of two strings are equal and do not exceed 10.
Output
Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10?-?9.
Sample test(s)
Input
++-+-+-+-+
Output
1.000000000000
Input
+-+-+-??
Output
0.500000000000
Input
+++??-
Output
0.000000000000
Note
For the first sample, both s1 and s2 will lead Dreamoon to finish at the same position ?+?1.
For the second sample, s1 will lead Dreamoon to finish at position 0, while there are four possibilites for s2: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5.
For the third sample, s2 could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position ?+?3 is 0.
立即学习“前端免费学习笔记（深入）”；
#include #include #include char s1[15],s2[15];int main(){ scanf("%s %s",s1,s2); int p1=0, m1=0; int p2=0, m2=0; int size=(int)strlen(s1); int wh=0; for(int i=0;i wh) { printf("0\n"); 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\n",double(ans)/double(all));} 立即学习“前端免费学习笔记（深入）”；
立即学习“前端免费学习笔记（深入）”；
立即学习“前端免费学习笔记（深入）”；
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
Input
2 2
Output
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}.
立即学习“前端免费学习笔记（深入）”；
#include #define ll long longconst int mod=1e9 + 7;int main(){ ll a,b; scanf("%I64d %I64d", &a, &b); if(b == 1) printf("0\n"); 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\n", (ans % mod + (temp % mod * b % mod)) % mod); }} 立即学习“前端免费学习笔记（深入）”；
立即学习“前端免费学习笔记（深入）”；
立即学习“前端免费学习笔记（深入）”；
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
51 2 3 5
Input
2 2
Output
222 4 6 2214 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 .
立即学习“前端免费学习笔记（深入）”；
#include int main(){ int n, k; scanf("%d %d", &n, &k); int a = 1, b = 2, c = 3, d = 5; printf("%d\n", (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\n",a, b, c, d); a += 6 * k; b += 6 * k; c += 6 * k; d += 6 * k; }} 立即学习“前端免费学习笔记（深入）”；
立即学习“前端免费学习笔记（深入）”；
