HDOJ 1865 1sting 13.04.21 周赛结题报告 （大数加法）

1sting

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2227 Accepted Submission(s): 875

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

128
解题思路：
一开始还以为要用组合数学来做了。。
后来发现原来是个斐波那契数列。
悲剧的是将200的字符串长度看成了20。。。 硬生生的把一道高精度加法当做了简单加法做。。
算法：
高精度计算。
代码：
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int a[201][101]={0};int main (){    int n,i,j;    a[1][100]=1;    a[2][100]=2;    for (i=3; i<=200; i++)    {        for (j=100; j>=0; j--)        {            a[i][j]+=a[i-1][j]+a[i-2][j];            if (a[i][j]>=10)            {                a[i][j-1]+=a[i][j]/10;                a[i][j]=a[i][j]%10;            }        }    }    cin>>n;    while(n--)    {        char str[210];        cin>>str;        int l=strlen(str);        int p;        for (j=0; j<=100; j++)            if (a[l][j]!=0)                break;        p=j;        for (j=p; j<=100; j++)            cout<<a[l][j];        cout<<endl;    }    return 0;}