【动态规划】[NOIP2003]数字游戏

题目

经典的环形DP 需要用到前缀和差分思想 f1数组去搞最大值 f2数组去搞最小值 f[i][j]表示 前i个数字 被分成了j份的 最大或者最小价值 枚举一个k表示那分出来的j份里面最后一份的开始位置 就很容易想出状态转移方程了

当然我还是看的题解

代码如下

#include<iostream>#include<cstdio>#include<cctype>    using namespace std;    #define in = read();    typedef long long ll;    typedef unsigned int ui;    const ll size = 500 + 10 ;        int n , m;        int a[size];        int maxx , minn = 0x3fff;        int sum[size] , f1[size][size] , f2[size][size];inline ll read(){        ll num = 0 , f = 1;    char ch = getchar();        while(!isdigit(ch)){                if(ch == '-')   f = -1;                ch = getchar();        }        while(isdigit(ch)){                num = num*10 + ch - '0';                ch = getchar();        }        return num*f;}inline void dp(int a[]){        for(register int i=1;i<=n;i++)                sum[i] = sum[i - 1] + a[i];        for(register int i=0;i<=n;i++)                for(register int j=0;j<=m;j++){                        f1[i][j] = 0;                        f2[i][j] = 0x3fff;                }        for(register int i=1;i<=n;i++)                f1[i][1] = f2[i][1] = (sum[i]%10 + 10)%10;        f1[0][0] = f2[0][0] = 1;        for(register int j=2;j<=m;j++)                for(register int i=j;i<=n;i++)                        for(register int k=j-1;k<=i-1;k++){                                f1[i][j] = max(f1[i][j] , f1[k][j - 1]*(((sum[i] - sum[k])%10 + 10)%10));                                f2[i][j] = min(f2[i][j] , f2[k][j - 1]*(((sum[i] - sum[k])%10 + 10)%10));                        }        maxx = max(maxx , f1[n][m]);        minn = min(minn , f2[n][m]);}int main(){        n in;   m in;        for(register int i=1;i<=n;i++){                a[i] in;                a[i + n] = a[i];        }        for(register int i=0;i<n;i++)                dp(a + i);        printf("%d\n%d\n" , minn , maxx);}

//COYG