By yusijia

Updated:

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解法一：

dp数组记录符合的所有可能
dp[n][1~3] 长度为n时，最后面是1~3个相同的

状态转移方程：
	dp[i][1] = 25 * (dp[i - 1][1] + dp[i - 1][2] + dp[i - 1][3]);
        dp[i][2] = dp[i - 1][1];
        dp[i][3] = dp[i - 1][2];


#include <iostream>
#include <cstdio>
#include <cstdlib>
#define MOD 1000000007

using namespace std;

long long dp[2005][4] = {0};

int main()
{
    dp[1][1] = 26;
    for(int i = 2; i <= 2000; i++){
        dp[i][1] = 25 * (dp[i - 1][1] + dp[i - 1][2] + dp[i - 1][3]) % MOD;
        dp[i][2] = dp[i - 1][1];
        dp[i][3] = dp[i - 1][2];
    }
    int t, n;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        printf("%d\n", int((dp[n][1] + dp[n][2] + dp[n][3]) % MOD));
    }
    return 0;
}


解法二：

状态转移方程：
dp[i] = 25 * (dp[i - 1] + dp[i - 2] + dp[i - 3]);

#include <cstdio>
#define MOD 1000000007

long long dp[2005] = {0};

int main()
{
    dp[1] = 26;
    dp[2] = 26 * 26;
    for (int i = 2; i <= 2000; ++i)
        dp[i]= 25 * (dp[i - 1] + dp[i - 2] + dp[i - 3]) % MOD;
    int t, n;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        printf("%d\n", int((dp[n] + dp[n - 1] + dp[n - 2]) % MOD));
    }
    return 0;
}
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