Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 208 Accepted Submission(s): 157
太水了
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013-9-15 0:04:49 4 File Name :G:\2013ACM练习\2013网络赛\2013成都网络赛\1003.cpp 5 ************************************************ */ 6 7 #pragma comment(linker, "/STACK:1024000000,1024000000") 8 #include <stdio.h> 9 #include <string.h> 10 #include <iostream> 11 #include <algorithm> 12 #include <vector> 13 #include <queue> 14 #include <set> 15 #include <map> 16 #include <string> 17 #include <math.h> 18 #include <stdlib.h> 19 #include <time.h> 20 using namespace std; 21 #define REP(I, N) for (int I=0;I<int(N);++I) 22 #define FOR(I, A, B) for (int I=int(A);I<int(B);++I) 23 #define DWN(I, B, A) for (int I=int(B-1);I>=int(A);--I) 24 #define REP_1(I, N) for (int I=1;I<=int(N);++I) 25 #define FOR_1(I, A, B) for (int I=int(A);I<=int(B);++I) 26 #define DWN_1(I, B, A) for (int I=int(B);I>=int(A);--I) 27 #define REP_C(I, N) for (int N____=int(N),I=0;I<N____;++I) 28 #define FOR_C(I, A, B) for (int B____=int(B),I=A;I<B____;++I) 29 #define DWN_C(I, B, A) for (int A____=int(A),I=B-1;I>=A____;--I) 30 #define REP_1_C(I, N) for (int N____=int(N),I=1;I<=N____;++I) 31 #define FOR_1_C(I, A, B) for (int B____=int(B),I=A;I<=B____;++I) 32 #define DWN_1_C(I, B, A) for (int A____=int(A),I=B;I>=A____;--I) 33 #define DO(N) while(N--) 34 #define DO_C(N) int N____ = N; while(N____--) 35 #define TO(i, a, b) int s_=a<b?1:-1,b_=b+s_;for(int i=a;i!=b_;i+=s_) 36 #define TO_1(i, a, b) int s_=a<b?1:-1,b_=b;for(int i=a;i!=b_;i+=s_) 37 #define SQZ(I, J, A, B) for (int I=int(A),J=int(B)-1;I<J;++I,--J) 38 #define SQZ_1(I, J, A, B) for (int I=int(A),J=int(B);I<=J;++I,--J) 39 40 41 int main() 42 { 43 //freopen("in.txt","r",stdin); 44 //freopen("out.txt","w",stdout); 45 int T; 46 scanf("%d",&T); 47 string str; 48 int iCase = 0; 49 while(T--) 50 { 51 iCase++; 52 cin>>str; 53 int n = str.length(); 54 if(n >= 4 && str[n-4] == 'd' && str[n-3] == 'e' && str[n-2] == 's' && str[n-1] == 'u') 55 { 56 str[n-4] = 'n'; 57 str[n-3] = 'a'; 58 str[n-2] = 'n'; 59 str[n-1] = 'o'; 60 str += "desu"; 61 } 62 else str += "nanodesu"; 63 printf("Case #%d: ",iCase); 64 cout<<str<<endl; 65 } 66 return 0; 67 }