ZZULIOJ1155:字符串比较多实例

tips:

1 不是一般的比较规则要跳出常规的眼光

2  观察找规律&＃xff0c;总体按照 a b c d等顺序&＃xff08;暂时先不看大小写&＃xff09;&＃xff0c;可以把单个字符看做是与 a 或 A的距离

3  按照ASII码 A

#include
#include
int main(){char buf1[10001], buf2[10001];int len1, len2,len;double num1, num2; //数字化表示的结果while (scanf("%s%s",buf1,buf2)!&＃61;EOF){len1 &＃61; strlen(buf1);len2 &＃61; strlen(buf2);len &＃61; len1 > len2 ? len1 : len2;for (int i &＃61; 0; i &＃61; &＃39;A&＃39;&&buf1[i] <&＃61; &＃39;Z&＃39;){num1 &＃61; (int)(buf1[i] - &＃39;A&＃39;)-0.5;}else if (buf1[i]>&＃61;&＃39;a&＃39;&&buf1[i]<&＃61;&＃39;z&＃39;){num1 &＃61; (int)(buf1[i] - &＃39;a&＃39;);}else{num1 &＃61; -1; //比较到了 &＃39;\0&＃39; ,结果应该是NO ,故第一个应大于第二个数字}if (buf2[i] >&＃61; &＃39;A&＃39;&&buf2[i] <&＃61; &＃39;Z&＃39;){num2 &＃61; (int)(buf2[i] - &＃39;A&＃39;) - 0.5;}else if (buf2[i] >&＃61; &＃39;a&＃39;&&buf2[i] <&＃61; &＃39;z&＃39;){num2 &＃61; (int)(buf2[i] - &＃39;a&＃39;);}else{num2 &＃61; -2; //比较到了 &＃39;\0&＃39;,结果应该是NO,故第二个应该小于第一个数字}if (num1num2){printf("NO\n");break;}}memset(buf1, &＃39;\0&＃39;, 10001);memset(buf2, &＃39;\0&＃39;, 10001);}
}