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贴代码,还是挺水的#include<cstdio> #include<iostream> using namespace std; int a[1005],dp[10005][35];//dp[i][j],i分钟时用了j次移动后得到的苹果数目 int main() { int i,j,n,m,t,Max = 0; scanf("%d %d",&n,&m); for(i = 1;i <= n;i ++) scanf("%d",&a[i]); for(i = 1;i <= n;i ++) { dp[i][0] = dp[i-1][0] + 2 - a[i];//只要是移动0次,则一定在苹果树1的位置 for(j = 1;j <= m;j ++)//若j为奇数则在苹果树2位置,否则在1位置 { /*之所以现在移动步数为j,是从j-1步移动过来的,或者是和上一步一样没有动*/ if(j % 2) dp[i][j] = max(dp[i-1][j-1],dp[i-1][j]) + a[i] - 1; else dp[i][j] = max(dp[i-1][j-1],dp[i-1][j]) + 2 - a[i]; } } for(i = 0;i <= m;i ++) Max = max(Max,dp[n][i]); printf("%d\n",Max); return 0; } Followed by:
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