HDU 5292 Pocket Cube

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结论题
问了学长怎么做的,学长告诉我这个结论,然后说,会玩魔方的人都知道啊…
这里写图片描述

//      whn6325689//        Mr.Phoebe//        http://blog.csdn.net/u013007900#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define eps 1e-9#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LLINF 1LL<<62#define speed std::ios::sync_with_stdio(false);typedef long long ll;typedef unsigned long long ull;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define CPY(x,y) memcpy(x,y,sizeof(x))#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define ls (idx<<1)#define rs (idx<<1|1)#define lson ls,l,mid#define rson rs,mid+1,r#define root 1,1,ntemplate<class T>inline bool read(T &n){    T x = 0, tmp = 1;    char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------const int MAXN=22;int sc[11][11]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,-1,0,0,0,0,0,0,0,0,-1,0,1,0,0,0,0,-1,0,1,0,0,0,0,0,-1,0,1,1,0,-1,0,0,0,0,0,1,0,-1,0,0,0,0,1,0,-1,0,0,0,0,0,0,0,0,-1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,};char str[MAXN];int main(){    freopen("data.txt","r",stdin);    int T,cas=1;    scanf("%d",&T);    while(T--)    {        getchar();        int ans=0;        for(int i=1;i<=8;i++)        {            gets(str);            int len=strlen(str);            for(int j=0;j<len;j++)                if(str[j]=='y' || str[j]=='w')                    ans+=sc[i][j];        }        printf("Case #%d: ",cas++);        if(ans%3==0)            puts("YES");        else            puts("NO");    }    return 0;} 
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