#include <stdio.h>
int main() {
int a, b = 1, i, j, aa[7][7], c[7];
scanf("%d\n", &a);
while (a > 0) {
for (j = 1; j <= 2; j++) {
scanf("%s", c);
aa[j][3] = c[0] - '0';
aa[j][4] = c[1] - '0';
aa[j][5] = c[2] - '0';
}
for (j = 3; j <= 5; j++) {
scanf("%s", c);
aa[j][1] = c[0] - '0';
aa[j][2] = c[1] - '0';
aa[j][3] = c[2] - '0';
aa[j][4] = c[3] - '0';
aa[j][5] = c[4] - '0';
aa[j][6] = c[5] - '0';
aa[j][7] = c[6] - '0';
}
for (j = 6; j <= 7; j++) {
scanf("%s", c);
aa[j][3] = c[0] - '0';
aa[j][4] = c[1] - '0';
aa[j][5] = c[2] - '0';
}
for (i = 3; i <= 5; i++) {
for (j = 1; j <= 2; j++) {
if (aa[i][j] == 1) {
if (aa[i][j + 1] == 1 && aa[i][j + 2] == 0)
b = 0;
if (aa[i + 1][j] == 1 && (aa[i + 2][j] == 0 || aa[i - 1][j] == 0))
b = 0;
}
}
for (j = 6; j <= 7; j++) {
if (aa[i][j] == 1) {
if (aa[i][j + 1] == 1 && aa[i][j + 2] == 0)
b = 0;
if (aa[i + 1][j] == 1 && (aa[i + 2][j] == 0 || aa[i - 1][j] == 0))
b = 0;
}
}
}
for (i = 1; i <= 7; i++) {
for (j = 3; j <= 5; j++) {
if (aa[i][j] == 1) {
if (aa[i][j + 1] == 1 && (aa[i][j + 2] == 0 || aa[i][j - 1] == 0))
b = 0;
if (aa[i + 1][j] == 1 && (aa[i + 2][j] == 0 || aa[i - 1][j] == 0))
b = 0;
}
}
}
a--;
if (b == 0)
printf("no\n");
else
printf("yes\n");
}
}
uj5u.com熱心網友回復：
不就是對每個空位置判斷其上下左右方向都不存在連續兩個棋子嗎？uj5u.com熱心網友回復：
主要是回圈的問題，因為上面兩排和下面兩排不是是從頭開始的
uj5u.com熱心網友回復：
你看看改成這樣行不行~~
#include <stdio.h>
int main() {
int i,j,n,flag;
char data[9][9]= {'\0'};
scanf("%d\n", &n);
while (n > 0) {
scanf("%s%s",data[1]+3,data[2]+3);
scanf("%s%s%s",data[3]+1,data[4]+1,data[5]+1);
scanf("%s%s",data[6]+3,data[7]+3);
flag=0;
for(i=1; i<8; i++) {
for(j=1; j<8; j++) {
if(data[i][j]!='1') continue; //當前位置沒棋子則跳過
if(data[i-1][j]=='1'&&data[i+1][j]=='0' //垂直方向：110
||data[i-1][j]=='0'&&data[i+1][j]=='1' //垂直方向：011
||data[i][j-1]=='1'&&data[i][j+1]=='0' //水平方向：110
||data[i][j-1]=='0'&&data[i][j+1]=='1') { //水平方向：011
flag=1; //出現以上4種情況表示有解
break;
}
}
if(flag==1) break;
}
n--;
if (flag == 1) printf("no\n");
else printf("yes\n");
}
return 0;
}
uj5u.com熱心網友回復：
謝謝大神
轉載請註明出處，本文鏈接：https://www.uj5u.com/houduan/270291.html
標籤：其它技術問題
上一篇：代碼錯哪了？
下一篇：數學建模