摘要： 点我看题目题意 ：中文题，不详述。思路 ： 状态转移方程 dp[ i ][ j ] = dp[i-1][k] + fabs(a[ i ][ j ]-a[i-1][k]) ;dp[i][j]代表的是在 i 时刻如果敲第j坐标上的地鼠需要的最小消耗。#include #include #include #include using namespace std ;int a[30][30] ;int dp[1000][510] ;const int INF = 99999999 ;int main(){ int N,K ; while(~scanf("%d %d",&N, 阅读全文

摘要： A. Nutstime limit per test：1 secondmemory limit per test:256 megabytesinput:standard inputoutput:standard outputYou haveanuts and lots of boxes. The boxes have a wonderful feature: if you putx(x ≥ 0)divisors (the spacial bars that can divide a box) to it, you get a box, divided intox + 1sections.You 阅读全文