首页 > 代码库 > hduoj 1865 1string 【大数】【菲波那切数列】
hduoj 1865 1string 【大数】【菲波那切数列】
1sting
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5310 Accepted Submission(s): 2030
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
【思路】1 先求题意,我们发现是s[1]=1,s[2]=2, 其余的 n>=3 时s[n]=s[n-1]+s[n-2];
2 此题最长的列为200,所以轻轻松松就超出int ,long long的范围,我们要改用数组a[205][102],来存储每一位上的数字,(前者[205]记录输入的1的个数,后者[102]记录结果的每一位)。
AC代码:
#include<stdio.h>#include<string.h>int a[205][102];//注意此处要比底下函数中的j的最大值开的大一点void count(){ int i,j,p,q; memset(a,0,sizeof(a));//数组清零 a[1][0]=1;a[2][0]=2; for(i=3;i<203;i++) //以下步骤模拟大数计算,初始化斐波那契数列 { p=q=0; for(j=0;j<=100;j++) { p=a[i-1][j]+a[i-2][j]+q; a[i][j]=p%10; q=p/10; }printf(" %d ",a[i][0]); }}int main(){ count(); int n,i,j,len; char s[205]; scanf("%d",&n); while(n--) { getchar(); scanf("%s",s); len=strlen(s); for(i=100;i>=0;i--)//找到数值的最后一位 if(a[len][i]!=0) break; for(j=i;j>=0;j--) // 注意上面的函数计算的值的数位是逆序的 printf("%d",a[len][j]); printf("\n"); } return 0;}
hduoj 1865 1string 【大数】【菲波那切数列】
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