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Problem Description
Fascinated with the computer games, Gabriel even forgets to study. Now she needs to finish her homework, and there is an easy problem:
f(n)=
She is required to calculate f(n) mod 2 for each given n. Can you help her?
Input
Multiple test cases. Each test case is an integer n(0≤n≤) in a single line.
Output
For each test case, output the answer of f(n)mod2.
//这道题一看数据范围就是到,不能用常规的递归来求解,找规律啊 啊啊啊啊啊
山东省第八届ACM省赛 I 题(Parity check)
#include <stdio.h>
#include<iostream>
#include <string.h>
#include<stdlib.h>
#include<string>
#include<cmath>
#include<algorithm>
#include<stdlib.h>
#define M 10005
using namespace std;
/*shandong ACM 8 I*/
//先按照普通方法找规律
/*int function(int n){
if(n==0){
return 0;
}
if(n==1){
return 1;
}
return function(n-1)+function(n-2);
}
int main(){
int n;
while(1){
scanf("%d",&n);
printf("%d\n",function(n)%2);
}
return 0;
}*/
//通过下面程序发现 if(n%3==0) return 0 else return 1
//接下来解决大数输入的问题 10的1000次方
int main(){
char c[1010];
while(scanf("%s",c)!=EOF){
int len=strlen(c);
int sum=0;
//判断每一位是否能被三{整除,若有一位不能被三整除,则 return 1
for(int i=0;i<len;i++) {
sum=(sum*10+(c[i]-'0'))%3; //实际上是从第一位对每一位取余数,当有余数时在对下一位取余数进位;
}
sum==0?printf("0\n"):printf("1\n");
return 0;
}
return 0;
}