前言

题目

大意是说，两个槽能够插在一起，并保证每一列的高度不高于3，保证最短长度。

思路

思路很简单，取短字符串遍历长字符串的每一个位置，纪录下位置，并取最短即可。

实现

//习题3-11 换抵挡装置
void caculate17() {char n1[1000], n2[1000];while (scanf("%s\n%s", n1, n2) != EOF) {char *min_arr, *max_arr;if (strlen(n1) < strlen(n2)) {min_arr = n1;max_arr = n2;}else {min_arr = n2;max_arr = n1;}int minLen = (int)strlen(min_arr);int maxLen = (int)strlen(max_arr);int min = minLen + maxLen;// 从最长数字的左边开始对齐for (int j = 0; j < maxLen; j++) {int j_temp = j;int result = 1;for (int i = minLen-1; i >= 0 && j_temp >= 0; i--) {int sum = (min_arr[i] - '0') + (max_arr[j_temp] - '0');if (sum > 3) {result = 0;break;}j_temp--;}if (result) {int temp = 0;if (j >= strlen(min_arr)-1) {temp = maxLen;}else {temp = minLen + maxLen - (j + 1);}if (temp < min) {min = temp;}}}if (strlen(min_arr) != strlen(max_arr)) {// 从最大数字的右边开始对齐for (int j = maxLen - minLen; j < maxLen; j++) {int j_temp = j;int result = 1;for (int i = 0; i < minLen && j_temp < maxLen; i++) {int sum = (min_arr[i] - '0') + (max_arr[j_temp] - '0');if (sum > 3) {result = 0;break;}j_temp++;}if (result) {int temp = 0;if (j == maxLen - minLen) {temp = maxLen;}else {temp = minLen + maxLen - (maxLen - j);}if (temp < min) {min = temp;}}}}printf("%d\n", min);}
}// 题目同上
void caculate18() {char n1[1000], n2[1000];while (scanf("%s\n%s", n1, n2) != EOF) {char *min_arr, *max_arr;if (strlen(n1) < strlen(n2)) {min_arr = n1;max_arr = n2;} else {min_arr = n2;max_arr = n1;}int maxLen = (int)strlen(max_arr);int minLen = (int)strlen(min_arr);int min = maxLen + minLen;int result = 0;for (int i = -minLen; i < maxLen + minLen; i++) {int ok = 1;// 判断for (int j = 0; j < minLen; j++) {// 重合部分if (i+j >= 0 && i+j < maxLen) {if ((max_arr[i + j] - '0') + (min_arr[j] - '0') > 3) {ok = 0;break;}}}// 短字符串在左边，部分重合if (ok && i < 0) {int temp = -i + maxLen; //-i代表左边的距离，移动距离则是重合部分if (temp < min) {min = temp;}}//短字符串在右边，部分重合if (ok && i > maxLen - minLen) {int temp = i + minLen; //这里的i则代表移动距离，没有包括重叠的部分if (temp < min) {min = temp;}}// 完全重合（最优解）if (ok && i >= 0 && i + minLen <= maxLen) {result = 1;break;}}if (result) {printf("%d\n", maxLen);}else {printf("%d\n", min);}}
}// 大神思路，简单高效
void caculate19() {char s1[128], s2[128];while (scanf("%s\n%s", s1, s2) == 2) {int max_len = strlen(s1), min_len = strlen(s2);int ret = min_len + max_len;// 从左到右for (int i = -min_len; i < max_len; i++) {int ok = 1;for (int j = 0; j < min_len && ok; j++) {if (i + j >= 0 && i + j < max_len) {ok &= s1[i+j] - '0'  + s2[j] - '0' <= 3;}}if (ok) {// 如果短字符串在左边或者重合的话，则为-i + max_len// 如果短字符串在右边的话，则为i + min_lenret = min(ret, max(max_len, i + min_len) - min(i, 0)); //太棒了!}}printf("%d\n", ret);}
}int main() {caculate18();return 0;
}

呃，我的实现是第一种，没ac，应该是只过了部分的测试用例，但我不知道怎么改了～