HDU 1865 （斐波拉切大数）

发布时间：2021-01-24 08:50 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): 5235????Accepted Submission(s): 1987 Problem Description You will be given a string which only contains ‘1’; You can merge

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.
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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

  
  
   
   3 1 11 22222

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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 ?
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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 250
char str[N];
int a[N][1100];   ///F[200]应该不会超过1100为的

void init()
{
    int i,j;
    memset(a,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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