1sting

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5021    Accepted Submission(s): 1864


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 3 1 11 11111
Sample Output 1 2 8
Author z.jt 這道題也是斐波那契數列，只是需要用字串把1的個數轉換為n而已。

#include<cstdio>
#include<cstring>
int main()
{
	int t,l,k;
	char a[201];
	int f[1001][211]={0};			//全部初始化為0 
	f[1][1]=1;
	f[2][1]=2;
	for(int i=3;i<=1000;i++)
	{
		for(int j=1;j<=210;j++)			//二維陣列，i表示第幾個數，j表示這個數的第幾位，
		{								// 然後對j進行類似大數相加的運算 
			f[i][j]+=f[i-1][j]+f[i-2][j];
			if(f[i][j]>9)
			{
				f[i][j]-=10;
				f[i][j+1]++;	
			}	
		}	
	}	
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s",a);
		l=strlen(a);
		k=100;
		for( ;f[l][k]==0;k--);			//開始全部初始化為零，直到遇到一個f[n][k]!=0,此時跳出迴圈，開始輸出 
		for( ;k>=1;k--)					//這種倒著輸出不同於之前的倒著輸出，此時不用考慮是否進位，只要不超出範圍即可 
			printf("%d",f[l][k]);
		printf("\n");
	}
	return 0;
}