c语言，检查密码的代码 循环到底哪出问题了啊为什么不对

Question

#include<stdio.h>
#include<string.h>
int main(){
int n,i,j,t,c,k,f;
char str[250]="";
scanf("%d",&n);
getchar();
for(i=0;i<n;i++){
gets(str);
t=strlen(str);
if(t<6) {
printf("Your password is tai duan le.\n");
continue;
}
else{
c=0,k=0,f=0;
for(i=0;i<t;i++){
if(str[i]>='0'&&str[i]<='9') {
c=1;
}else if(str[i]>='A'&&str[i]<='Z'||str[i]>='a'&&str[i]<='z'){
k=1;
}else if(str[i]=='.') {
f=1;
}
}
if(f==0) printf("Your password is tai luan le.\n");
else if(c==0) printf("Your password needs zi mu.\n");
else if(k==0) printf("Your password needs shu zi.\n");
else if(c==1&&k==1) printf("Your password is wan mei.\n");

}

}

return 0;

}

Answer 1

第二个循环不要使用i，应该用j：

字母和数字的判断反了，应该这样子：

密码里面没有'.'符号时的提示错了，应该不是提示“太短了”，而应该提示没有'.'符号

#include<stdio.h>
#include<string.h>
int main() {
    int n, i, j, t, c, k, f;
    char str[250] = "";
    scanf("%d", &n);
    getchar();
    for (i = 0; i < n; i++) {
        gets(str);
        t = strlen(str);
        if (t < 6) {
            printf("Your password is tai duan le.\n");
            continue;
        }
        else {
            c = 0, k = 0, f = 0;
            for (j = 0; j < t; j++) {
                if (str[j] >= '0' && str[j] <= '9') {
                    c = 1;
                }
                else if (str[j] >= 'A' && str[j] <= 'Z' || str[j] >= 'a' && str[j] <= 'z') {
                    k = 1;
                }
                else if (str[j] == '.') {
                    f = 1;
                }
            }
            if (f == 0) printf("Your password needs dian.\n");
            else if (k == 0) printf("Your password needs zi mu.\n");
            else if (c == 0) printf("Your password needs shu zi.\n");
            else printf("Your password is wan mei.\n");

        }
    }
    return 0;
}