題目鏈接

思路自西瓜and大佬博客：https://www.cnblogs.com/henry-1202/p/10590327.html#_label3

數據範圍小 可直接dp

f[i][j][a][b] 表示 i位置上是j i-1上是a i-2上是b

狀態轉移是由i-1轉移過來的，所以就必須還要一個i-3 所以就多加上一個循環

最主要就是轉移過程中要枚舉每種情況 然後排除掉

大佬的博客用了map和string簡化了枚舉的過程

不過他說只有六種情況 我布吉島為啥只有六種 我寫出了八種qaq

map<int, string> mp;
bool check(int a, int
 
 b, int c, int d) {
    if (mp[a] + mp[b] + mp[c] == "AGC") return 0;
    if (mp[a] + mp[c] + mp[b] == "AGC") return 0;
    if (mp[a] + mp[b] + mp[d] == "AGC") return 0;
    if (mp[a] + mp[c] + mp[d] == "AGC") return 0;
    if (mp[b] + mp[a] + mp[c] == "AGC") return 0;
    if (mp[b] + mp[c] + mp[d] == "
 
AGC") return 0;
    if (mp[b] + mp[d] + mp[c] == "AGC") return 0;
    if (mp[c] + mp[b] + mp[d] == "AGC") return 0;
    return 1;
}

mp[0]=‘E‘就是來處理剛開始未到3個字符時候的情況 不過枚舉j也就是第i個位置的字符就從1開始了 所以之後的不會受到mp[0]的影響了

#include <cstdio>
#include <algorithm>
#include <string>
#include 
 
<map>
using namespace std;

map<int, string> mp;
long long f[110][5][5][5];
const long long mod = 1e9 + 7;

bool check(int a, int b, int c, int d) {
    if (mp[a] + mp[b] + mp[c] == "AGC") return 0;
    if (mp[a] + mp[c] + mp[b] == "AGC") return 0;
    if (mp[a] + mp[b] + mp[d] == "AGC") return 0;
    if (mp[a] + mp[c] + mp[d] == "AGC") return 0;
    if (mp[b] + mp[a] + mp[c] == "AGC") return 0;
    if (mp[b] + mp[c] + mp[d] == "AGC") return 0;
    if (mp[b] + mp[d] + mp[c] == "AGC") return 0;
    if (mp[c] + mp[b] + mp[d] == "AGC") return 0;
    return 1;
}

int main() {
    int n;
    scanf("%d", &n);
    f[0][0][0][0] = 1;
    mp[0] = ‘E‘, mp[1] = ‘A‘, mp[2] = ‘C‘, mp[3] = ‘G‘, mp[4] = ‘T‘;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= 4; j++) 
            for (int a = 0; a <= 4; a++)
                for (int b = 0; b <= 4; b++)
                    for (int l = 0; l <= 4; l++)
                        if (check(l, b, a, j)) 
                            f[i][j][a][b] = (f[i][j][a][b] + f[i-1][a][b][l]) % mod;
    long long ans = 0;
    for (int i = 1; i <= 4; i++) 
        for (int j = 1; j <= 4; j++)
            for (int k = 1; k <= 4; k++)
                ans = (ans + f[n][i][j][k]) % mod;
    printf("%lld\n", ans);
    return 0;
}

AtCoder Beginner Contest 122 D - We Like AGC(DP)