1sting+大fibonacci数列求和

1sting

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2237 Accepted Submission(s): 883



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
The output contain n lines, each line output the number of result you can get .

[align=left]Sample Input[/align]

3
1
11
11111

[align=left]Sample Output[/align]

1
2
8

/*思路：这题就是一个大fibinacci数列求和，*/
#include<iostream>
using namespace std;
char f[205][102];
void cal(char a[],char b[],int n){
strrev(a);
strrev(b);
int lena=strlen(a);
int lenb=strlen(b);
int i,max=(lena>lenb)?lena:lenb;
for(i=lena;i<max;i++)
a[i]='0';
a[i]='\0';
for(i=lenb;i<max;i++)
b[i]='0';
b[i]='\0';
int sum=0,num=0,k=0;
for(i=0;i<max;i++){
sum=a[i]-'0'+b[i]-'0'+num;
f
[k++]=sum%10+'0';
num=sum/10;
}
while(num)
{
f
[k++]=num%10+'0';
num/=10;
}
f
[k]='\0';
strrev(f
);
}
void fun()
{
memcpy(f[1],"1\0",2);
memcpy(f[2],"2\0",2);
for(int i=3;i<=200;i++){
char a[102],b[102];
memcpy(a,f[i-1],102);

memcpy(b,f[i-2],102);
cal(a,b,i);
}

}

int main()
{
int t;
cin>>t;
fun();
while(t--)
{
char a[201];
cin>>a;
int len=strlen(a);
int lenf=strlen(f[len]);
for(int i=0;i<lenf;i++)
if(f[len][i])
break;
for(;i<lenf;i++)
cout<<f[len][i];
cout<<endl;
}
return 0;
}