HDU 1865 （斐波拉切大数）

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1sting

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5235 Accepted Submission(s): 1987

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

311111111

Sample Output

Author

z.jt

Source

2008杭电集训队选拔赛——热身赛

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题意：

所给由1组成的串，每相邻的两个1可以变成一个2，求共有多少种不同的变法；

推出几组，你就可以发现是斐波拉切数列，即F[i]=F[i-1]+F[i-2];这里的i就是所给串的长度；

#include<iostream>#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>typedef long long ll;using namespace std;#define INF 0x3f3f3f3f#define N 250char str[N];int a[N][1100];   ///F[200]应该不会超过1100为的void init(){    int i,j;    memset(a,0,sizeof(a));    a[1][0]=1;    a[2][0]=2;    for(i=3;i<=201;i++)    {        for(j=0;j<1100;j++)        {            a[i][j] += a[i-1][j]+a[i-2][j];            if(a[i][j]>=10)            {                a[i][j+1] =a[i][j+1]+a[i][j]/10;    ///刚开始把这里两个等式位置颠倒了，一直不进位，搞得很无奈啊。。。傻!!!                a[i][j]=a[i][j]%10;            }        }    }}int main(){    int n,i,len,index;    init();    scanf("%d",&n);    while(n--)    {        scanf("%s",str);        len=strlen(str);        for(i=1099;i>=0;i--)        {            if(a[len][i]!=0)            {                index=i;                break;            }        }        for(i=index;i>=0;i--)        {            printf("%d",a[len][i]);        }        printf("\n");    }    return 0;}

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