题意：给你一个数字n,有两种操作：减1或乘2，问最多经过几次操作能变成m；
随后发篇随笔普及下memset函数的初始化问题。自己也是涨了好多姿势。
代码
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #define INF 0x7fffffff; 6 using namespace std; 7 int DP[20100], vis[20100]; 8 int dp(int n, int m) 9 { 10 if(n <= 0) return INF; 11 if(n >= m) {DP[n] = n-m; return DP[n];} 12 if(!vis[n]) 13 { 14 vis[n] = 1; 15 //cout << n << "+++" << endl; 16 DP[n] = min(dp(n-1, m), dp(n*2, m))+1; 17 } 18 return DP[n]; 19 } 20 int main() 21 { 22 int n, m; 23 while(~scanf("%d%d", &n, &m)){ 24 memset(DP, 0x3f3f3f3f, sizeof(DP)); 25 memset(vis, 0, sizeof(vis)); 26 //cout << DP[0] << endl; 27 printf("%d\n", dp(n, m)); 28 } 29 return 0; 30 }