简单DP + 高精

洛谷P1255、P2437.

思路相同，注意何时进位以及数组最大长度！

#include <stdio.h>

short dp[5020][2000] = { 0 };

void f(int n) {
    dp[0][0] = 0;
    dp[1][0] = 1;
    dp[2][0] = 2;
    
    int flag = 0;
    for (int i = 3; i <= n; i++) {
        for (int j = 0; j <= flag; j++) {
            dp[i][j] += dp[i - 1][j] + dp[i - 2][j];
            if (dp[i][j] >= 10) {
                dp[i][j] %= 10;
                dp[i][j + 1]++;
                if (j == flag)flag++;
            }
        }
    }

    for (flag; flag >= 0; flag--) {
        printf("%d", dp[n][flag]);
    }
}

int main() {
    int n;
    scanf("%d", &n);
    f(n);
    return 0;
}

#include <stdio.h>

short dp[1010][500] = { 0 };

void f(int m, int n) {
    int des = 1 + n - m;
    int flag = 0;
    dp[1][0] = 0;
    dp[2][0] = 1;
    dp[3][0] = 2;
    for (int i = 4; i <= des; i++) {
        for (int j = 0; j <= flag; j++) {
            dp[i][j] += dp[i - 1][j] + dp[i - 2][j];
            if (dp[i][j] >= 10) {
                dp[i][j] %= 10;
                dp[i][j + 1]++;
                if (j == flag)flag++;
            }
        }
    }
    for (flag; flag >= 0; flag--) {
        printf("%d", dp[des][flag]);
    }
}

int main() {
    int m, n;
    scanf("%d%d", &m, &n);
    f(m, n);
    return 0;
}