题解 | 二维斐波那契数列

#include <inttypes.h>
#include <stdio.h>
#include <string.h>

int main() {
    int n,m;
    const int MOD = 1000000007;
    scanf("%d %d",&n,&m);
    long long a[n+1][m+1];
    memset(a, 0, sizeof(a));
    a[1][1]=1;
    for(int i=2;i<=n;i++){
        a[i][1]=(a[i-1][1])%MOD;
    }
    for(int j=2;j<=m;j++){
        a[1][j]=(a[1][j-1])%MOD;
    }
       for (int i = 2; i <= n; i++) {
        for (int j = 2; j <= m; j++) {
            a[i][j] = (a[i - 1][j] + a[i][j - 1]) % MOD;
        }
    }
    printf("%lld\n",a[n][m]%MOD);
}