ACdream 1667 调皮的数一 (大数+dp)

题目链接：
​​​ACdream 1667​​

题解：
很显然是dp题。
当时状态可以由左边的跑道，右边的跑道，跑道不变这三种状态转换而来。
设dp[i][j]表示人跑了i 步跑到j 跑道的方案数。
那么，dp[i][j]=dp[i−1][j+1]+dp[i−1][j]+dp[i−1][j−1].

注意：这道题大数啊…套个C++大数类，或者上Java吧…我一般不写Java…

WA代码：

/*
* this code is made by LzyRapx
* Problem: 1667
* Verdict: Wrong Answer
* Submission Date: 2017-06-04 16:24:13
* Time: 32MS
* Memory: 13564KB
*/
#include <bits/stdc++.h>
using namespace std;
long long  dp[1234][1234];
int main()
{
    int n;
    dp[0][0]=1;
    for(int i=1;i<=1001;i++)
    {
        for(int j=1;j<=1001;j++)
        {
            dp[i][j] = dp[i-1][j+1]+(dp[i-1][j]+dp[i-1][j-1]);
        }
    }
    while(cin>>n)
    {   
        cout<<dp[n+1][1]<<endl;
    }
    return 0;

大数C++代码：

/*
* this code is made by LzyRapx
* Problem: 1667
* Verdict: Accepted
* Submission Date: 2017-06-04 19:56:33
* Time: 1252MS
* Memory: 3332KB
*/
#include <bits/stdc++.h>
using namespace std;
const int N=1100,M=1009;
const int MX = 150;
const int MAXN = 9999;
const int DLEN = 4;
class Big {
public:
    int a[MX], len;
    Big(const int b = 0) {
        int c, d = b;
        len = 0;
        memset(a, 0, sizeof(a));
        while(d > MAXN) {
            c = d - (d / (MAXN + 1)) * (MAXN + 1);
            d = d / (MAXN + 1);
            a[len++] = c;
        }
        a[len++] = d;
    }
    Big(const char *s) {
        int t, k, index, L, i;
        memset(a, 0, sizeof(a));
        L = strlen(s);
        len = L / DLEN;
        if(L % DLEN) len++;
        index = 0;
        for(i = L - 1; i >= 0; i -= DLEN) {
            t = 0;
            k = i - DLEN + 1;
            if(k < 0) k = 0;
            for(int j = k; j <= i; j++) {
                t = t * 10 + s[j] - '0';
            }
            a[index++] = t;
        }
    }
    Big operator/(const int &b)const {  //大数除以整数
        Big ret;
        int i, down = 0;
        for(i = len - 1; i >= 0; i--) {
            ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
            down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
        }
        ret.len = len;
        while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
        return ret;
    }
    bool operator>(const Big &T)const {
        int ln;
        if(len > T.len) return true;
        else if(len == T.len) {
            ln = len - 1;
            while(a[ln] == T.a[ln] && ln >= 0) ln--;
            if(ln >= 0 && a[ln] > T.a[ln]) return true;
            else return false;
        } else return false;
    }
    Big operator+(const Big &T)const {
        Big t(*this);
        int i, big;
        big = T.len > len ? T.len : len;
        for(i = 0; i < big; i++) {
            t.a[i] += T.a[i];
            if(t.a[i] > MAXN) {
                t.a[i + 1]++;
                t.a[i] -= MAXN + 1;
            }
        }
        if(t.a[big] != 0) t.len = big + 1;
        else t.len = big;
        return t;
    }
    Big operator-(const Big &T)const {
        int i, j, big;
        bool flag;
        Big t1, t2;
        if(*this > T) {
            t1 = *this;
            t2 = T;
            flag = 0;
        } else {
            t1 = T;
            t2 = *this;
            flag = 1;
        }
        big = t1.len;
        for(i = 0; i < big; i++) {
            if(t1.a[i] < t2.a[i]) {
                j = i + 1;
                while(t1.a[j] == 0) j++;
                t1.a[j--]--;
                while(j > i) t1.a[j--] += MAXN;
                t1.a[i] += MAXN + 1 - t2.a[i];
            } else t1.a[i] -= t2.a[i];
        }
        t1.len = big;
        while(t1.a[t1.len - 1] == 0 && t1.len > 1) {
            t1.len--;
            big--;
        }
        if(flag) t1.a[big - 1] = 0 - t1.a[big - 1];
        return t1;
    }
    int operator%(const int &b)const {   //大数对余数取模
        int i, d = 0;
        for(i = len - 1; i >= 0; i--) {
            d = ((d * (MAXN + 1)) % b + a[i]) % b;
        }
        return d;
    }
    Big operator*(const Big &T) const {
        Big ret;
        int i, j, up, temp, temp1;
        for(i = 0; i < len; i++) {
            up = 0;
            for(j = 0; j < T.len; j++) {
                temp = a[i] * T.a[j] + ret.a[i + j] + up;
                if(temp > MAXN) {
                    temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
                    up = temp / (MAXN + 1);
                    ret.a[i + j] = temp1;
                } else {
                    up = 0;
                    ret.a[i + j] = temp;
                }
            }
            if(up != 0) {
                ret.a[i + j] = up;
            }
        }
        ret.len = i + j;
        while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
        return ret;
    }
    void print() {
        printf("%d", a[len - 1]);
        for(int i = len - 2; i >= 0; i--) printf("%04d", a[i]);
        puts("");
    }
};

int main()
{
    Big one("1");
    Big zero("0");
    //注意：内存不够，要用滚动数组 
    Big dp[3][M];  
    dp[1][1]=one;  
    dp[1][2]=one;
    dp[0][1]=dp[1][1];
    for(int i=2;i<M;i++)
    {
        for(int j=1;j<M;j++)
        {
            if(j>1)dp[2][j] = dp[1][j - 1]; //左边 
            else dp[2][j]=zero;
            dp[2][j] = dp[2][j] + dp[1][j]; //不变 
            if(j+1<1005) dp[2][j] = dp[2][j] + dp[1][j + 1]; //r右边 
        }
        for(int j=1;j<=M;j++) dp[1][j]=dp[2][j];

        dp[0][i]=dp[1][1];  
    }
    int n;
    while(cin>>n)
    {
        dp[0][n].print();
    }
    return 0;
}