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超大空间 超长时间 超长代码 水过~开头0数据是必须的#include<stdio.h> #include<string.h> int a[11][11][11][11][11][11][11] ; int bi[8] ; int map(char c){ switch(c){ case 'A':case 'B': case 'C':return 2;break ; case 'D':case 'E': case 'F':return 3;break ; case 'G':case 'H': case 'I':return 4;break ; case 'J':case 'K': case 'L':return 5;break ; case 'M':case 'N': case 'O':return 6;break ; case 'P':case 'R': case 'S':return 7;break ; case 'T':case 'U': case 'V':return 8;break ; case 'W':case 'X': case 'Y':return 9;break ; } return -1 ; } int maked(char c){ if(c == '-') return -1 ; else if(c >= '0' && c <= '9') return c - '0' ; else{ if(c == 'Z' || c == 'Q') return -1 ; return map(c) ; } } int print(){ int ok = 0 ; for(int i = 0 ; i < 10 ; i++) for(int j = 0 ; j < 10 ; j++) for(int k = 0 ; k < 10 ; k++) for(int l = 0 ; l < 10 ; l++) for(int m = 0 ; m < 10 ; m++) for(int n = 0 ; n < 10 ; n++) for(int o = 0 ; o < 10 ; o++) if(a[i][j][k][l][m][n][o] > 1){ printf("%d%d%d-%d%d%d%d %d\n" ,i,j,k,l,m,n,o,a[i][j][k][l][m][n][o]) ; ok = 1 ; } return ok ; } int main() { char t[200] ; int n ; scanf("%d" , &n) ; while(n--){ scanf("%s" , t) ; int k = 1 , temp ; memset(bi , 0 , sizeof(bi)) ; //memset(t , 0 , sizeof(t)) ; for(int i = 0 ; i < strlen(t) ; i++) if((temp = maked(t[i])) >= 0){ bi[k++] = temp ; } a[bi[1]][bi[2]][bi[3]][bi[4]][bi[5]][bi[6]][bi[7]]++ ; } //printf("%d\n" , print()) ; if(!print()) printf("No duplicates. \n") ; return 0 ; } Followed by: Post your reply here: |
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