hdu 5455

来源：互联网发布：hadoop python api 编辑：程序博客网时间：2024/06/04 17:47
水题，对字符串简单处理一下就好了。
#include <cstdio>#include <stack>#include <set>#include <iostream>#include <string>#include <vector>#include <queue>#include <list>#include <functional>#include <cstring>#include <algorithm>#include <cctype>#include <string>#include <map>#include <iomanip>#include <cmath>#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define MP(a, b) make_pair(a, b)#define MS(arr, num) memset(arr, num, sizeof(arr))#define PB push_back#define F first#define S second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define lrt rt << 1#define rrt rt << 1|1#define root 1,n,1#define BitCount(x) __builtin_popcount(x)#define BitCountll(x) __builtin_popcountll(x)#define LeftPos(x) 32 - __builtin_clz(x) - 1#define LeftPosll(x) 64 - __builtin_clzll(x) - 1const double PI = acos(-1.0);const int INF = 1e7;using namespace std;const double eps = 1e-5;const int MAXN = 300 + 10;const int MOD = 1000007;const int N=1000100;const int MAX=log2(N*1.0);typedef pair<int, int> pii;typedef pair<int, string> pis;int n,m,ans,vis[N],v;char s[N];int slove(char *c){    int i=0,j,k,len=strlen(c);    while(i<len) {        if (c[i++]=='c') {            k=0;            while(i<len && c[i]=='f') i++,k++;            if (k<2) {                if (i==len) return k;                else {                    return -1;                }            }            else if (i==len) break;            else  {                ans++;            }        }    }    return k;}int main(){    int i,j,T,K=1;    cin>>T;    getchar();    while(T--)    {        ans=0;        int flag=0;        gets(s);  // 可能有空格        for (i=0;s[i];i++) if(s[i]!='c' && s[i]!='f') break;//有c 和 f 之外的字符的话直接输出-1        if (s[i]) {            printf("Case #%d: -1\n",K++);            continue ;        }        for (i=0;s[i];i++) if (s[i]=='c') {            break;        }        v=i;        if (!s[i]) {            if (i<=2) ans=1;            else if (i&1) ans=i/2+1;            else ans=i/2;        }        else {            int k=slove(s+i);            if (k<0 || k+v<2) ans=-1;  处理头尾            else  {                ans++;            }        }        printf("Case #%d: %d\n",K++,ans);    }}
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