浙江大学PAT上机题解析之1014. 福尔摩斯的约会 (20)
1014. 福尔摩斯的约会 (20)#include <cstdio>#include <cstring>#include <cstdlib>int main(){ char str[4][61]; memset(str,0,sizeof(str)); scanf("%s%s%s%s",str[0],str[1],str[2],str[3]); int len1 = strlen(str[0]); int len2 = strlen(str[1]); int len3 = strlen(str[2]); int len4 = strlen(str[3]); int i=0,j=0; for (;i<len1&&j<len2;i++,j++) { if (str[0][i]==str[1][j] && str[0][i]>='A' &&str[0][i]<='G') { switch(str[0][i]) {case 'A':printf("MON ");break;case 'B':printf("TUE ");break;case 'C':printf("WED ");break;case 'D':printf("THU ");break;case 'E':printf("FRI ");break;case 'F':printf("SAT ");break;case 'G':printf("SUN ");break; } break; } } for (i++,j++;i<len1&&j<len2;i++,j++) { if (str[0][i]==str[1][j] && (str[0][i]>='A' &&str[0][i]<='N' || str[0][i]>='0'&&str[0][i]<='9')) { int t=0; if (str[0][i]>='A'&&str[0][i]<='N') t = str[0][i] - 'A' +10; else t = str[0][i]-'0'; printf("%d%d:",t/10,t%10); break; } } for (i=0,j=0;i<len3&&j<len4;i++,j++) { if (str[2][i]==str[3][j] && (str[2][i]>='A' && str[2][i]<='Z'||str[2][i]>='a'&&str[2][i]<='z')) { printf("%d%d\n",i/10,i%10); break; } } //system("pause");}
