hd1865 1sting

发布时间：2020-12-14 02:27:02 所属栏目：大数据来源：网络整理

导读：1sting Time Limit: 5000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4154????Accepted Submission(s): 1562 Problem Description You will be given a string which only contains ‘1’; You can merge tw

1sting

Time Limit: 5000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4154????Accepted Submission(s): 1562

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

嗯，题目大意就是一串以1组成的字符串，每两个1可以替换成一个2，问一个串可以有多少种写法。

多写几个算一下，就可以看出来是Fibonacci数列，只不过这个不能直接用一位数组打表存数据了，字符串最长200，需要用大数加法了

#include<cstdio>
#include<cstring>
char s[300];
int va[300][300];
void dabiao()
{
	int i,j;
	memset(va,sizeof(va));
	va[1][0]=1;
	va[2][0]=2;
	for(i=3;i<201;++i)
	{
		for(j=0;j<=200;++j)
		{
			va[i][j]=va[i][j]+va[i-1][j]+va[i-2][j];
			if(va[i][j]>=10)
			{
				va[i][j]-=10;
				va[i][j+1]+=1;
			}
		}
	}
}
int main()
{
	int n,i,len;
	scanf("%d",&n);
	dabiao();
	while(n--)
	{
		scanf("%s",s);
		len=strlen(s);
		i=200;
		while(va[len][i]==0)i--;
		for(;i>=0;i--)printf("%d",va[len][i]);
		printf("n");
	}
	return 0;
}

（编辑：李大同）

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