题解 | 二维斐波那契数列

#include <stdio.h>


int main() {
    int n,m;
    scanf("%d %d",&n,&m);
    long anm[n][m];
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j ++){
            if(i == 0 || j == 0){
               anm[i][j] = 1;
            }
            else{
                anm[i][j] = (anm[i - 1][j] % (1000000000 + 7) + anm[i][j - 1]% (1000000000 + 7))% (1000000000 + 7);
            }
        }
    }
    int x = anm[n - 1][m - 1];
    printf("%d",x);
    return 0;
}