2013暑假集训B组训练赛第二场 - E Queue at the School
Description
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 ton, 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 tseconds.
Input
The first line contains two integers n and t (1 ≤ n, t ≤ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
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".
Sample Input
5 1
BGGBG
GBGGB
5 2
BGGBG
GGBGB
4 1
GGGB
GGGB
依然是水题,犯二了,开始还弄了几个数组
#include <cstdio> int x, y; int main() { while(~scanf("%d%d",&x,&y)) { while(x >= 0&& y>=0) { if(x>=2 && y >= 2) { x -= 2; y -= 2; } else if(x==1 && y >= 12) { x -= 1; y -= 12; } else if(x== 0 && y>=22) { y -= 22; } else { puts("Ciel"); break; } if(y>=22) { y -= 22; } else if(y>=12 && x >=1) { x -= 1; y -= 12; } else if(x>= 2 && y >=2) { x -= 2; y -= 2; } else { puts("Hanako"); break; } } } }
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