问题 B: Divisible by 3 规律题

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问题 B: Divisible by 3时间限制: 1 Sec  内存限制: 128 MB提交: 291  解决: 86[提交][状态][讨论版]题目描述Let's consider a sequence 1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, 123456789, 12345678910, 1234567891011, ....Write a program that determines how many elements of this sequence among first n are divisible by three.输入One positive integer n (1 ≤ n ≤ 231 - 1).输出You need to print one number you've found.样例输入4样例输出2提示

来源: http://125.221.232.253/JudgeOnline/problem.php?cid=1204&pid=1

/*规律: f(n)%3=[(f(n-1)%3)*10+n%3]%3而数字是以3K+1 3K+2 3(K+1)循环出现的3K+1=1(mod 3)那么 对于下一个序列数就有 1*10+3K+2 = 0(mod 3)最后一个 0*10+3(K+1) = 0(mod 3)所以循环节 1 0 0N个数里一共有N/3组和N%3个余项*/#include <cstdio>int main(void){    int N;    while(~scanf("%d",&N))        printf("%d\n",(N/3)*2+(N%3?N%3-1:0));    return 0;}
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