Lightoj 1136 Division by 3 (idea title)

http://www.lightoj.com/volume_showproblem.php?problem=1136

Start to solve two times with the same, calculate a will find a lot of solutions. --> instead of parsing the sequence structure.

Analysis found:

A number of numbers that consist of three consecutive integers together and bound to be divisible by 3. (X+x+1+x+2=3x+3=3 (x+1))

Thus there are:

1. The number of 3rd K in the title must be divisible by three.

2. The 3rd K + 1 number in the title, its 2nd number to the last number of the sum must be divisible by 3, plus the first number, that is 1, can not be 3 whole.

3. The 3rd K + 2 number in the title, the sum of the 3rd number to the last digit must be divisible by 3, plus the first to second number, which is 1 + 2 = 3, and divisible by 3.

So the 1~k can be divisible by 3 with 2k/3+ (k%3==2?1:0).

Complete code:

01./*0.008s,1088kb*/
.  
#include <cstdio>  
04.typedef unsigned int ll;  
06.ll Calc (ll k)  
07.{  
Return    K/3 * 2 + (k 3 = 2 1:0);  
.  
11.int Main ()  
12.{    int T, cas = 0;    ll A, B;    scanf ("%d", &t);  
While    (t--)  
.    {        scanf ("%u%u", &a, &b);        printf ("Case%d:%u\n", ++cas, Calc (b)-Calc (A-1));  
.    return 0;  
22.}

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