*大数求和加上斐波拉契

Problem O
Time Limit: 1000 MS Memory Limit: 32 MB 64bit IO Format: %I64d
Submitted: 218 Accepted: 41
[Submit][Status][Web Board]
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

3
1
11
22222

Sample Output

1
2
8

大数求和加上斐波拉契

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main()
{
    int i,j,cnt,r,n,len1,len2,k,m,temp,carry,t,x,flag;

    char c,a[1000],b[1000],d[1000];
    scanf("%d",&n);
    getchar();
    for(i=0; i<n; i++)
    {
        cnt=0;
        while(1)
        {
            scanf("%c",&c);
            if(c=='1')
                cnt++;
            else
                break;
        }
         sprintf(a,"%d",1);
           sprintf(b,2);
        for(j=3; j<=cnt; j++)
        {
          len1=strlen(a);
          len2=strlen(b);
         x=len1>len2?len1:len2;
          carry=0;
            t=x;
          memset(d,'0',t);
          d[t+1]='';

          for(k=len1-1,m=len2-1;k>=0||m>=0;k--,m--,t--)
          {
              temp=carry;
              if(k>=0)temp+=a[k]-'0';
              if(m>=0)temp+=b[m]-'0';
              if(temp>=10)
              {
                  d[t]=temp-10+'0';
                  carry=1;
              }
              else
              {
                  d[t]=temp+'0';
                  carry=0;
              }

          }
            d[t]=carry+'0';
            strcpy(a,b);
            strcpy(b,d);
        }
        flag=0;
        if(cnt==1)
            printf("1");

        else if(cnt==2)
            printf("2");
        else
          for(j=0;j<=x;j++)
          {
              if(d[j]!='0')
                flag=1;
              if(flag==1)
                printf("%c",d[j]);

          }
          printf("n");
    }
    return 0;
}