HDOJ 题目18651sting（大数 斐波那契）
发布时间：2020-12-14 03:28:21 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): 3191????Accepted Submission(s): 1223 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)Total Submission(s): 3191????Accepted Submission(s): 1223
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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Author
z.jt
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Source
2008杭电集训队选拔赛——热身赛
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? ac代码 #include<stdio.h>
#include<string.h>
int a[1001][10000];
char * add(char a[],char b[],char c[])
{
int len1,len2,i,j,t[10000],max,k=0;
len1=strlen(a);
len2=strlen(b);
i=len1-1;
j=len2-1;
memset(t,sizeof(t));
while(i>=0||j>=0)
{
if(i<0&&j>=0)
t[k]+=b[j]-'0';
else
if(j<0&&i>=0)
t[k]+=a[i]-'0';
else
{
t[k]+=a[i]-'0'+b[j]-'0';
}
k++;
t[k]+=t[k-1]/10;
t[k-1]%=10;
if(t[k])
max=k;
else
max=k-1;
i--;
j--;
}
for(i=max;i>=0;i--)
c[max-i]=t[i]+'0';
c[max+1]=' ';
return c;
}
void fun()
{
int i;
strcpy(a[1],"1");
strcpy(a[2],"2");
for(i=3;i<=210;i++)
{
add(a[i-1],a[i-2],a[i]);
}
}
int main()
{
int n;
fun();
scanf("%d",&n);
while(n--)
{
int num;
char s[210];
scanf("%s",s);
num=strlen(s);
printf("%sn",a[num]);
}
}
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