黄龙洞传送阵
题目描述
为了更好地能将中华五千年文化融入到科技教育中，他们决定到远古时期去走一走。他们知道杭州黄龙洞有一座传
送阵，可以传送到古代。那天，他们来到了黄龙洞，发现传送阵上有四个按钮，每个按钮上有个算式，说出2个正
整数a和b,然后代入这个算式，计算出来的结果如果能被2整除，那么他们将回到夏朝，如果能被3整除，他们将回
到汉朝，如果能被5整除，他们将回到唐朝，能被7整除，他们将回到元朝，如果上述情况都不行，传送送将不起作
用，如果能被多个整除，将按上述顺序第一个的生效。现在知道这四个算式为
（1）2a+3b-7ab
(2)5a+9
(3)7a-4b
(4)ab输入
输入3个正整数，分别为a b和算式号输出
输出分2行，第1行输出算式的值，第2行根据传送朝代输出拼音，
夏朝 汉朝 唐朝 元朝 未能传送
xia han tang yuan wei样例输入
1 3 2样例输出
14
xia题解：
L1之题，水题无题解。（只要N个IF）
源代码：
#include<stdio.h>
int main() {
int a,b,c,s=0;
scanf("%d%d%d",&a,&b,&c);
if(c==1) {
s=2*a+3*b-7*a*b;
if(s%2==0) {
printf("%d\n",s);
printf("xia\n");
} else if(s%3==0) {
printf("%d\n",s);
printf("han\n");
} else if(s%5==0) {
printf("%d\n",s);
printf("tang\n");
} else if(s%7==0) {
printf("%d\n",s);
printf("yuan\n");
} else if((s%2==0&&s%5==0)||(s%2==0&&s%7==0)||(s%2==0&&s%3==0)||(s%3==0&&s%5==0)||(s%3==0&&s%7==0)||(s%5==0&&s%7==0)||(s%2==0&&s%3==0&&s%4==0)||(s%2==0&&s%3==0&&s%5==0)||(s%2==0&&s%4==0&&s%5==0)||(s%3==0&&s%4==0&&s%5==0)) {
printf("%d\n",s);
printf("xia\n");
} else if(s%2!=0&&s%3!=0&&s%5!=0&&s%7!=0) {
printf("%d\n",s);
printf("wei\n");
}
}
s=0;
if(c==2) {
s=5*a+9;
if(s%2==0) {
printf("%d\n",s);
printf("xia\n");
} else if(s%3==0) {
printf("%d\n",s);
printf("han\n");
} else if(s%5==0) {
printf("%d\n",s);
printf("tang\n");
} else if(s%7==0) {
printf("%d\n",s);
printf("yuan\n");
} else if((s%2==0&&s%5==0)||(s%2==0&&s%7==0)||(s%2==0&&s%3==0)||(s%3==0&&s%5==0)||(s%3==0&&s%7==0)||(s%5==0&&s%7==0)||(s%2==0&&s%3==0&&s%4==0)||(s%2==0&&s%3==0&&s%5==0)||(s%2==0&&s%4==0&&s%5==0)||(s%3==0&&s%4==0&&s%5==0)) {
printf("%d\n",s);
printf("xia\n");
} else if(s%2!=0&&s%3!=0&&s%5!=0&&s%7!=0) {
printf("%d\n",s);
printf("wei\n");
}
}
s=0;
if(c==3) {
s=7*a-4*b;
if(s%2==0) {
printf("%d\n",s);
printf("xia\n");
} else if(s%3==0) {
printf("%d\n",s);
printf("han\n");
} else if(s%5==0) {
printf("%d\n",s);
printf("tang\n");
} else if(s%7==0) {
printf("%d\n",s);
printf("yuan\n");
} else if((s%2==0&&s%5==0)||(s%2==0&&s%7==0)||(s%2==0&&s%3==0)||(s%3==0&&s%5==0)||(s%3==0&&s%7==0)||(s%5==0&&s%7==0)||(s%2==0&&s%3==0&&s%4==0)||(s%2==0&&s%3==0&&s%5==0)||(s%2==0&&s%4==0&&s%5==0)||(s%3==0&&s%4==0&&s%5==0)) {
printf("%d\n",s);
printf("xia\n");
} else if(s%2!=0&&s%3!=0&&s%5!=0&&s%7!=0) {
printf("%d\n",s);
printf("wei\n");
}
}
s=0;
if(c==4) {
s=a*b;
if(s%2==0) {
printf("%d\n",s);
printf("xia\n");
} else if(s%3==0) {
printf("%d\n",s);
printf("han\n");
} else if(s%5==0) {
printf("%d\n",s);
printf("tang\n");
} else if(s%7==0) {
printf("%d\n",s);
printf("yuan\n");
} else if((s%2==0&&s%5==0)||(s%2==0&&s%7==0)||(s%2==0&&s%3==0)||(s%3==0&&s%5==0)||(s%3==0&&s%7==0)||(s%5==0&&s%7==0)||(s%2==0&&s%3==0&&s%4==0)||(s%2==0&&s%3==0&&s%5==0)||(s%2==0&&s%4==0&&s%5==0)||(s%3==0&&s%4==0&&s%5==0)) {
printf("%d\n",s);
printf("xia\n");
} else if(s%2!=0&&s%3!=0&&s%5!=0&&s%7!=0) {
printf("%d\n",s);
printf("wei\n");
}
}
return 0;
}AC
