Fibonacci

题目链接

Problem Description
Fibonacci numbers are well-known as follow:

 
Now given an integer N, please find out whether N can be represented as the sum of several Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers.

Input
Multiple test cases, the first line is an integer T (T<=10000), indicating the number of test cases.
Each test case is a line with an integer N (1<=N<=10^9).

Output
One line per case. If the answer don’t exist, output “-1” (without quotes). Otherwise, your answer should be formatted as “N=f1+f2+…+fn”. N indicates the given number and f1, f2, … , fn indicating the Fibonacci numbers in ascending order. If there are multiple ways, you can output any of them.
 
Sample Input
4
5
6
7
100

Sample Output
5=5
6=1+5
7=2+5
100=3+8+89

Source
“浪潮杯”山东省第七届ACM大学生程序设计竞赛

 

代码:

#include<bits/stdc++.h>
using namespace std;
//一般就是定义为最大的变量 + 10,开始定义为visited[44]错了多少次,我吉吉都付不清了......
long long int a,b,k,flag,f[50],fx[50],visited[50];
void dfs(long long int ans){
    //由于只需要输出一组,所以定义一个flag变量,如果已经进行完一组,直接返回
    if(flag)
        return;
    if(ans == b){
        flag = 1;
        printf("%lld=",b);
        for(int i = k - 1;i >= 0;i--){
            if(i == k - 1)
                printf("%lld",fx[i]);
            else
                printf("+%lld",fx[i]);
        }
        printf("\n");
        return;
    }
    for(int i = 44;i >= 1;i--){//从大往小找,节省时间
        //题目要求不能连续,即必须相差一个,因此!visited[i - 1] && !visited[i + 1]也必须要判断
        if(ans + f[i] <= b && !visited[i] && !visited[i - 1] && !visited[i + 1]){
            visited[i] = true;
            fx[k++] = f[i];
            dfs(ans + f[i]);
            if(flag)//如果进行完一组,直接返回,不用向下递归
                return;
            visited[i] = false;
            k--;//回溯
        }
    }
    return;
}
int main(){
    f[1] = 1,f[2] = 2;
    for(int i = 3;i <= 44;i++)
        f[i] = f[i - 1] + f[i - 2];//先进行打表
    while(~scanf("%lld",&a)){
        while(a--){
            memset(fx,0,sizeof(fx));
            memset(visited,0,sizeof(visited));
            scanf("%lld",&b);
            flag = 0,k = 0;
            dfs(0);
            if(!flag)
                printf("-1\n");
        }
    }
    return 0;
}