uva 769(构造)
2015-04-23 23:57
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题意:给出一个n*n的网格,每个网格上有一张纸,网格是从左到右从上到下编号1~n * n,然后一个人最初站在格子1上,然后左右上下任意走到有纸片的格子上走k1步(k >= 2 * n),拿走一些纸片,保证一定不会拿走此时的落脚点上的纸,然后继续走k2步(k2 >= 2 * n),最后只剩下一张纸就是此时站的位置,输出每次的步数k和拿走的纸的网格编号。
题解:这个是有规律的,一行一行的删除,和初始点1距离为偶数的格子在走奇数步中一定走不到,距离为奇数的格子在走偶数步中一定走不到,最后一行一张一张的删除,可以剩下一个。
#include <stdio.h>
int main() {
int t, n;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
int step = 2 * n;
for (int i = 0; i < n - 1; i++) {
int lim = n * (i + 1);
printf("%d", step++);
for (int j = 2 + (i * n); j <= lim; j += 2)
printf(" %d", j);
printf("\n");
printf("%d", step++);
for (int j = 1 + (i * n); j <= lim; j += 2)
printf(" %d", j);
printf("\n");
}
step--;
int temp = n * (n - 1) + 1;
for (int i = 0; i < n - 1; i++)
printf("%d %d\n", step += 2, temp++);
if (t)
printf("\n");
}
return 0;
}
题解:这个是有规律的,一行一行的删除,和初始点1距离为偶数的格子在走奇数步中一定走不到,距离为奇数的格子在走偶数步中一定走不到,最后一行一张一张的删除,可以剩下一个。
#include <stdio.h>
int main() {
int t, n;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
int step = 2 * n;
for (int i = 0; i < n - 1; i++) {
int lim = n * (i + 1);
printf("%d", step++);
for (int j = 2 + (i * n); j <= lim; j += 2)
printf(" %d", j);
printf("\n");
printf("%d", step++);
for (int j = 1 + (i * n); j <= lim; j += 2)
printf(" %d", j);
printf("\n");
}
step--;
int temp = n * (n - 1) + 1;
for (int i = 0; i < n - 1; i++)
printf("%d %d\n", step += 2, temp++);
if (t)
printf("\n");
}
return 0;
}
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