这个我记得是有更优解的，不过还是明天发吧，明天想一想，看看能不能想起来
#include<string.h>
int main()
{char a[201] = { 0 };char b[201] = { 0 };scanf("%s %s", a, b);int na = strlen(a);int nb = strlen(b);int i = 0, j = 0;if (na >= nb){for(i=0;i<na;i++){int flag = 1;for (j = 0; j < nb; j++){if (a[i+j] != b[j]){flag = 0; break; }}if (flag){printf("%s is substring of %s\n", b, a);return 0;}}}else{for (i = 0; i < nb; i++){int flag = 1;for (j = 0; j < na; j++){if (b[i + j] != a[j]){flag = 0; break;}}if (flag){printf("%s is substring of %s\n", a, b);return 0;}}}printf("No substring\n");return 0;
}
这串代码我自己测是没问题的，可是检测过不去。
#include<string.h>
#include<ctype.h>
int main()
{char ch[101] = { 0 };gets(ch);char x[100] = { 0 };gets(x);char y[100] = { 0 };gets(y);int nx = strlen(x);int ny = strlen(y);int n = strlen(ch);ch[n] = 32;int i = 0, j = 0, m = 0;char a[100][100] = { 0 };//分单词for (i = 0; i < n; i++){int nm = 0;while (!isspace(ch[i])){a[m][nm++] = ch[i++];}m++;}//查看是否一样，且替换for (i = 0; i < m; i++){//查看是否为待替换单词for (j = 0; j < nx; j++){if (a[i][j] != x[j])break;}//如果是，替换if (j == nx){for (j = 0; j < ny; j++){a[i][j] = y[j];}a[i][j] = 0;//这个在替换后单词后面那个元素赋值0，防止原单词长，到时候打印出没有替换的那一部分}printf("%s ", &a[i][0]);}}
找到了
，不过不具体说了，可以自己对比一下
#include<string.h>
#include<ctype.h>
int main()
{char ch[102] = { 0 };gets(ch);char x[100] = { 0 };char y[100] = { 0 };scanf("%s %s",x, y);int nx = strlen(x);int ny = strlen(y);int n = strlen(ch);ch[n] = 32;int i = 0, j = 0, m = 0;char a[101][101] = { 0 };//分单词for (i = 0; i < n; i++){j = 0;while (!isspace(ch[i])){a[m][j++] = ch[i++];}m++;}//查看是否一样，且替换for (i = 0; i < m; i++){//查看是否为待替换单词for (j = 0; j < nx; j++){//这里要是，nx==1，且a和x的第一个一样，后面x没有了，但下面替换条件还是满足的if (a[i][j] != x[j])break;}//如果是，替换if (a[i][j]==0&&j==nx)//所以这里加了个看看a[i][j]有没有到最后{for (j = 0; j < ny; j++){a[i][j] = y[j];}a[i][j] = 0;//这个在替换后单词后面那个元素赋值0，防止原单词长，到时候打印出没有替换的那一部分}printf("%s ", &a[i][0]);}}
