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| #include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
long long num[150],dp_1[150],dp_2[150];//_1放最大值,_2放最小值
int n,m;
inline int mod(long long nnn){//mod函数
int w = int(floor(double(nnn)/10));
return nnn - w*10;
}
int main(){
scanf("%d %d",&n,&m);
for(int i = 1;i<=n;i++){//断环为链的准备
scanf("%lld",&(num[i]));
num[i+n] = num[i];
}
for(int i = 1;i<=2*n;i++)//前缀和的处理
num[i] = num[i-1]+num[i];
//正经dp
long long maxn = -1,minn = 0x3f3f3f3f;
for(int s = 1;s<=n;s++){//枚举起点
memset(dp_1,0,sizeof(dp_1));//清零dp数组
memset(dp_2,0,sizeof(dp_2));
for(int j = 0;j<m;j++){//板子由少到多
for(int i = s;i<=s+n;i++){//上一个插板子的地方
if(j == 0){//j==0的时候的处理，其实也可以叫初始化
dp_1[i] = dp_2[i] = mod(num[s+n]-num[i]);
continue;
}
long long maxtmp = -1,mintmp = 0x3f3f3f3f;//对所有可能下一状态的遍历，并取最大或最小值
for(int x = i+1;x<=s+n-j-1;x++){
maxtmp = max(maxtmp,dp_1[x]*mod(num[x]-num[i]));
mintmp = min(mintmp,dp_2[x]*mod(num[x]-num[i]));
}
dp_1[i] = maxtmp;//取最大最小值
dp_2[i] = mintmp;
}
}
maxn = max(dp_1[s],maxn);//对于起点不同的最大最小值进行更新
minn = min(dp_2[s],minn);
}
printf("%lld\n%lld\n",minn,maxn);
return 0;
}
|
