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During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
GGGB
题目的大意是给你一堆汉子妹子然后给定一个时间然后在这个时间内，如果汉子在妹子前面，汉子就得跟妹子换位置这道题首先得注意可以换位置的时间因为这关乎汉子能最后换到哪里去的问题接下来再搜索需要换位子的汉子，然后换成功一次后就得跳过该人 进入下一个人的判断 代码如下 #include #include using namespace std;int main(){ int n,t; cin>>n>>t; char str[105]; char s; cin>>str; int i,j; for(j=0;j
A. Lucky Division time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number n is almost lucky. Input The single line contains an integer n (1 ≤ n ≤ 1000) — the number that needs to be checked. Output In the only line print "YES" (without the quotes), if number n is almost lucky. Otherwise, print "NO" (without the quotes). Sample test(s) input 47 output YES input 16 output YES input 78 output NO Note Note that all lucky numbers are almost lucky as any number is evenly divisible by itself. In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4. 唔这道题的大意是让你搜索一个范围的的接近幸运数，我们把只含4或7的数称之为幸运数，而接近幸运数指的一个能被幸运数整除的数字是故范围更大，这道题是纯暴力做法因为给定的范围很小，所以我很偷懒的直接把这个范围内的幸运数枚举了出来然后再将输入的数依次跟幸运数去取余如果等于0这个数就是接近幸运数，当然如果给定的范围很大则用循环找出这个大范围内的所有幸运数并存在数组里然后再用N去对这个数组里的数依次取余只要有一个为0那就是接近幸运数了 代码如下~ #include using namespace std;int main(){ int n,m,t=0,c=0,a; cin>>n; {if(n%4==0||n%7==0||n%44==0||n%77==0||n%74==0||n%47==0||n%444==0||n%447==0||n%474==0||n%477==0||n%744==0||n%747==0||n%774==0||n%777==0) cout<<"YES"< using namespace std;int main(){ int i,n,a[100000],b[100000],u=0; cin>>n; for(i=0;i >a[i]>>b[i]; } for(i=0;i #include #include #include using namespace std;int main(){ char a[1005],b[1005],c[1005]; int len1,len2,len3; int i; cin>>a>>b>>c; len1=strlen(a); len2=strlen(b); len3=strlen(c); strcat(a,b); sort(a,a+len1+len2); sort(c,c+len3); if(strcmp(a,c)==0) cout<<"YES"<
Consider the first test sample. Manao can fail his first push and push the wrong button. In this case he will already be able to guess the right one with his second push. And his third push will push the second right button. Thus, in the worst-case scenario he will only need 3 pushes.
这道题是一道简单的数学题，一开始把它理解为错排了后来认真看了一下发现与错排还是有区别的 大意是给你一串密码 然后问你最悲催的得按多少下那个键盘你才能过 注意每按错一个密码就得重新开始 而且还是一个一个的按。 唔这道题的数据是有规律的首先1对应1 2对应3 3对应7 我自己再模拟了4 5 6 分别为14 25 41乍看之下木有规律 但是不难发现 无论如何最后一次都是对的 都得按那最后一个正确的密码 所以可以把每个数先减去1 得到0 2 6 13 24 40 唔 这有啥规律呢 后一个减去前一个试试看呗 不难发现 出现了2 4 7 11 16 这些数 再减一次发现为2 3 4 5 成为了一个等差数列 是故如果把0 2 6 13 24 这些数设为a[i]那么a[i]-a[i-1]（i从1开始）的差就是一个首项为2 公差为1的前N项数列的和的通项公式为(i*i+i)/2+1, 最后递归得到a[n-1]再加上1就是答案了 代码如下 #include using namespace std;int main(){ int a[100005]; int n; cin>>n; a[0]=0; a[1]=2; int i; for(i=1;i
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