1sting+大fibonacci数列求和

最新推荐文章于 2024-10-11 18:29:31 发布

李逝水

最新推荐文章于 2024-10-11 18:29:31 发布

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本文介绍了一个基于Fibonacci数列求和的问题，通过字符串操作和递推算法来解决由一串数字'1'组成的特定组合问题。文章提供了一段C++代码实现，该算法能够计算出所有可能的不同结果数量。

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1sting

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2237 Accepted Submission(s): 883

Problem Description

You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given 1111 , you can get 1111, 121, 112,211,22. Now, your work is to find the total number of result you can get.

Input

The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200.

Output

The output contain n lines, each line output the number of result you can get .

Sample Input

Sample Output

/*思路：这题就是一个大fibinacci数列求和，*/
#include<iostream>
using namespace std;
char f[205][102];
void cal(char a[],char b[],int n){
	strrev(a);
	strrev(b);
	int lena=strlen(a);
	int lenb=strlen(b);
	int i,max=(lena>lenb)?lena:lenb;
	for(i=lena;i<max;i++)
		a[i]='0';
	a[i]='\0';
	for(i=lenb;i<max;i++)
		b[i]='0';
	b[i]='\0';
	int sum=0,num=0,k=0;
	for(i=0;i<max;i++){
		sum=a[i]-'0'+b[i]-'0'+num;
		f[n][k++]=sum%10+'0';
		num=sum/10;
	}
	while(num)
	{
		f[n][k++]=num%10+'0';
		num/=10;
	}
	f[n][k]='\0';
	strrev(f[n]);
}
void fun()
{
	memcpy(f[1],"1\0",2);
	memcpy(f[2],"2\0",2);
	for(int i=3;i<=200;i++){
		char a[102],b[102];
		memcpy(a,f[i-1],102);
		
		memcpy(b,f[i-2],102);
		cal(a,b,i);
	}
	
}

int main()
{
	int t;
	cin>>t;
	fun();
	while(t--)
	{
		char a[201];
		cin>>a;
		int len=strlen(a);
		int lenf=strlen(f[len]);
		for(int i=0;i<lenf;i++)
			if(f[len][i])
				break;
			for(;i<lenf;i++)
				cout<<f[len][i];
			cout<<endl;
	}
	return 0;
}