pku2033Alphacode(zjgsu1216) 递推

用递推，

对当前考虑的第i位进行考虑，表达成i-1或i-2的关系式

今天做这两道题目花了很多时间，不想写了，代码核心部分看看很容易懂的

代码如下：

 

#include<stdio.h>
#include<string.h>
char s[10005];
__int64 dp[10005];
void process()
{
    int len=strlen(s),i,temp;
    dp[0]=1;
    for(i=1;i<len;i++)
    {
        temp=(s[i-1]-'0')*10+s[i]-'0';
        if(s[i]=='0')
        {    if(i==1)
                dp[i]=1;
            else
            dp[i]=dp[i-2];
        }
        else if(s[i-1]=='0'||temp>26)
            dp[i]=dp[i-1];
        else if(temp>9&&temp<=26)
        {    if(i==1)
                dp[i]=2;
            else
            dp[i]=dp[i-1]+dp[i-2];
        }
    }
    printf("%I64d\n",dp[len-1]);

}
int main()
{
    while(scanf("%s",s),s[0]!='0')
    {
        process();
    }
    return 0;
}

#include<stdio.h>
#include<string.h>
char s[502];
__int64 d;
__int64 dp[502];
void process()
{
    int i,temp,flag=0;
    dp[0]=1;
    for(i=1;i<strlen(s);i++)
    {
        temp=(s[i-1]-'0')*10+s[i]-'0';
        if(s[i-1]=='0'||temp>=26)
            dp[i]=dp[i-1];
        else if(temp<=25&&temp>9)
        {    
            if(i==1)
                dp[i]=2;
            else
                dp[i]=dp[i-1]+dp[i-2];
        }
        if(dp[i]>d)
        {
            printf("NO\n");
            flag=1;break;
        }
    }
    if(flag==0)
        printf("ACCEPT\n");
}
int main()
{
    while(scanf("%s%I64d",s,&d)!=EOF)
    {
        process();
    }
    return 0;
}