HDOJ 1865 1sting（大数斐波那契数列）
发布时间：2020-12-14 02:33:34 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): 4024????Accepted Submission(s): 1497 Problem Description You will be given a string which only contains ‘1’; You can merge tw
1stingTime Limit: 5000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
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
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Sample Output
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和南阳OJ上的655题《光棍的yy》一样，代码题解相同，代码如下：
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#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,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;
}
}
}
int main()
{
count();
int n,i,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;
}
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