Codeforces 412A Poster(贪心)
2014-04-20 12:33
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题目链接:Codeforces 412A Poster
题目大意:公司要为大标题上色,给出标题的长度,接着给出标题,移动梯子的代价是比较大的,所以要求找到最小的移动方案为标题的每个字上色。
解题思路:其实只要保证就近原则,梯子的移动次数就是最小的,首先看梯子更靠近左边还是右边,将它移动至较为靠近的一边,然后逐个输出即可。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 105;
int main () {
int n, k;
char s
;
scanf("%d%d%s", &n, &k, s);
if (k <= n - k) {
for (int i = 1; i < k; i++)
printf("LEFT\n");
printf("PRINT %c\n", s[0]);
for (int i = 1; i < n; i++) {
printf("RIGHT\n");
printf("PRINT %c\n", s[i]);
}
} else {
for (int i = 0; i < n - k; i++)
printf("RIGHT\n");
printf("PRINT %c\n", s[n-1]);
for (int i = n-2; i >= 0; i--) {
printf("LEFT\n");
printf("PRINT %c\n", s[i]);
}
}
return 0;
}
题目大意:公司要为大标题上色,给出标题的长度,接着给出标题,移动梯子的代价是比较大的,所以要求找到最小的移动方案为标题的每个字上色。
解题思路:其实只要保证就近原则,梯子的移动次数就是最小的,首先看梯子更靠近左边还是右边,将它移动至较为靠近的一边,然后逐个输出即可。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 105;
int main () {
int n, k;
char s
;
scanf("%d%d%s", &n, &k, s);
if (k <= n - k) {
for (int i = 1; i < k; i++)
printf("LEFT\n");
printf("PRINT %c\n", s[0]);
for (int i = 1; i < n; i++) {
printf("RIGHT\n");
printf("PRINT %c\n", s[i]);
}
} else {
for (int i = 0; i < n - k; i++)
printf("RIGHT\n");
printf("PRINT %c\n", s[n-1]);
for (int i = n-2; i >= 0; i--) {
printf("LEFT\n");
printf("PRINT %c\n", s[i]);
}
}
return 0;
}
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