摘要：核心代码： for(i = 1; i <= n; i++){ pre[i][i] = a[i]; for(j = i + 1; j <= n; j++){ pre[i][j] = a[j] * pre[i][j - 1] % mod; printf("pre[%d][%d] = %d ",i,j,p 阅读全文

摘要：样例输入 10 3 18 7 14 10 12 23 41 16 24 样例输出 6 #include <stdio.h> int a[1005], dp[1005]; int main() { int i, j, l, max; int n; scanf("%d",&n); for(i = 1; 阅读全文