#include#include#includeusing namespace std;int main(){char s[1001];while(gets(s)){if(strcmp(s,"zoj")==0||strcmp(s,"zj")==0){cout
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#include#include#includeusing namespace std;int main(){char s[1001];while(gets(s)){if(strcmp(s,"zoj")==0||strcmp(s,"zj")==0){cout
#include
#include
#include
using namespace std;
int main()
{
char s[1001];
while(gets(s))
{
if(strcmp(s,"zoj")==0||strcmp(s,"zj")==0)
{
cout
#include#include#includeusing namespace std;int main(){char s[1001];while(gets(s)){if(strcmp(s,"zoj")==0||strcmp(s,"zj")==0){cout
貌似你连样例都还不能通过哈,觉得你应该至少把样例通过了再说,如果给你说明了你还做题干嘛.
oozoojoooo
这个样例貌似有问题,我这样觉得只这样,对有问题.
1,2,3规则都不符合~不晓得是不是我理解有误哈,
不过据说应该这样做:
1)确定有且只有一个z和j.
2)确定z在j前面.
3)判断zoj和zj的情况.
4)判断xzojx的情况..即前后的o的个数相等,其实判断zoj是判断这个类型的逻辑子集.
5)判断azbjc,看a,b,c是否相等...zj的判断是判断此类的子集.
6)最后据说这判断azbjc 是否有a*b=c这个条件,至于为什么,不知道.
综合发现第三步可以省略掉,貌似就这几个步骤哈~
#include
#include #include #include #include #include #include
#include #include #include
#include #include #include
实在看不懂#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;int cases;int main() {double cl = clock();scanf(%d,
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