【杭电oj】1865 - 1sting（大数递推）

发布时间：2021-03-01 18:23:18 所属栏目：大数据来源：网络整理

导读：点击打开题目 1sting Time Limit: 5000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4973????Accepted Submission(s): 1842 Problem Description You will be given a string which only contains ‘1’; You

点击打开题目

1sting

Time Limit: 5000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4973????Accepted Submission(s): 1842

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

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Sample Output

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Author z.jt ?
Source 2008杭电集训队选拔赛——热身赛

观察了一下，就是大数的斐波那契数列。

代码如下：

#include <cstdio>
#include <cstring>
int main()
{
	int k,u,n;
	char num[211];
	int fi[1011][151] = {0};
	fi[0][1] = 1;
	fi[1][1] = 1;
	for (int i = 2 ; i <= 1000 ; i++)
	{
		for (int j = 1 ; j <= 150 ; j++)
		{
			fi[i][j] += fi[i-1][j] + fi[i-2][j];
			if (fi[i][j] >= 10)
			{
				fi[i][j] -= 10;
				fi[i][j+1]++;
			}
		}
	}
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%s",num);
		n = strlen(num);
		k = 150;
		for ( ; fi[n][k] == 0 ; k--);
		for ( ; k >= 1 ; k--)
			printf ("%d",fi[n][k]);
		printf ("n");
	}
	return 0;
}

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