zoj 3131 Digital Clock

神马也不说了，打表很强大 = =。。。

 

直接打表，不符合电子表显示的数都不要，每个数里存从000000到这个数中符合题意的数。

 

如果是第一天到第二天，分成两次求和即可。

 

#include <stdio.h>#include <stdlib.h>#include <iostream>#include <string.h>using namespace std;int ans;int num[300000];int com(int x,int y,int z){return x*10000+y*100+z;}int init(){int i,sum = 0;memset(num,0,sizeof(num));for(i=0; i<=235959; i++){if( i%100 >= 60 || i%10000 >= 6000 )num[i] = -1;if( num[i] == 0 )if(i % 3 == 0 ){sum++;num[i] = sum;}elsenum[i] = sum;}}int process(int s1,int s2,int s3,int e1,int e2,int e3){int ans = 0,i;int num1 = com(s1,s2,s3);int num2 = com(e1,e2,e3);ans = num[num2] - num[num1];if( num1 % 3 == 0 ) ans++;return ans;}int main(){int ncases;char str[20];int s1,s2,s3,e1,e2,e3,len,ans,i,num1,num2;init();scanf("%d",&ncases);while( ncases-- ){scanf("%s",str);len = strlen(str);for(i=0; i<len; i++)if( str[i] == ':' )str[i] = ' ';sscanf(str,"%d %d %d",&s1,&s2,&s3);scanf("%s",str);len = strlen(str);for(i=0; i<len; i++)if( str[i] == ':' )str[i] = ' ';sscanf(str,"%d %d %d",&e1,&e2,&e3);num1 = com(s1,s2,s3);num2 = com(e1,e2,e3);if( num2 > num1 )ans = process(s1,s2,s3,e1,e2,e3);else{ans = process(s1,s2,s3,23,59,59);ans += process(0,0,0,e1,e2,e3);}printf("%d/n",ans);}return 0;}