int main() {
int s, b;
int hist[26] = { 0 };
int hist2[26] = { 0 };
char char1, char2;
printf("Hello Netta, enter the letters you would like us to repeat ending with $:\n");
scanf("%c", &char2);
while (char2 != '$') {
if (char2 >= 'A' && char2 <= 'Z')
char2 = char2 + 32;
int char3 = char2 - 'a';
hist2[char3]++;
scanf("%c", &char2);
if (char2 < 0)
break;
}
printf("How many times would you like to loop?\n");
if (!scanf("%d", &s))
return 0;
printf("Enter the string you would like to be checked ending with $:\n");
scanf("%c", &char1);
if (char1 >= 'A' && char1 <= 'Z')
char1 = char1 + 32;
while (char1 != '$' && char1 > 0) {
int char3 = char1 - 'a';
hist[char3]++;
scanf("%c", &char1);
}
for (int i = 0; i < 26; i++) {
if (hist[i] > s * hist2[i]) {
printf("Not enough letters\n");
b = 0;
break;
} else {
b = 1;
}
}
if (b)
printf("Congratulations! you have enough letters to create your song and win the Eurovision!\n");
return 0;
}
//所以基本上这是我大学的一个主页,他们要求我们做程序输入是char和一个循环,它将它与另外一个输入进行比较,你可以循环每个字母多少次(不需要检查输入是,但循环数int
答案 0 :(得分:0)
你的程序有很多问题:
scanf()
返回成功转化的次数。将返回值与程序中的1
进行比较,而不是测试0
,这对于"%c"
来说绝对不会发生。此外,如果流位于文件末尾,则不会修改char2
。
在索引到数组之前,您必须检查char2
是否为字母,否则您可能会超出数组边界并且具有未定义的行为。
演示很重要:使用适当的缩进和间距使程序可读。
包含必要的标题,例如<stdio.h>
以下是改进版本:
#include <stdio.h>
int main() {
int hist[26] = { 0 };
int hist2[26] = { 0 };
int s;
char c;
printf("Hello Netta, enter the letters you would like us to repeat ending with $:\n");
while ((scanf("%c", &c) == 1 && c != '$') {
if (c >= 'A' && c <= 'Z')
hist2[c - 'A']++;
else if (c >= 'a' && c <= 'z')
hist2[c - 'a']++;
}
printf("How many times would you like to loop?\n");
if (scanf("%d", &s) != 1)
return 1;
printf("Enter the string you would like to be checked ending with $:\n");
while (scanf("%c", &c) == 1 && c != '$') {
if (c >= 'A' && c <= 'Z')
hist[c - 'A']++;
else if (c >= 'a' && c <= 'z')
hist[c - 'a']++;
}
for (int i = 0; i < 26; i++) {
if (hist[i] > s * hist2[i]) {
printf("Not enough letters\n");
break;
}
}
if (i == 26)
printf("Congratulations! you have enough letters to create your song and win the Eurovision!\n");
return 0;
}