Integer partitioning (interval DP) dynamic programming

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For two integers n, m, we need to add M-1 in N and divide n into m segments to find out the maximum product of this m segment. (Number of digits of 1<=n<=1e19,m<=n)

Exercises

1, set Dp[i][j][k] is the maximum product of k in the interval [i,j] multiplication number.

2, note that when merging the 22 merge into Dp[i][j][0], it is directly merged into an integer without multiplication.

3, 22 when merging into Dp[i][j][k] (k!=0), we use multiplication to merge them together, and there is no omission.

Code:

#include <cstdio> #include <cstring> #include <iostream> #include <sstream> #include < algorithm> #include <vector> #include <bitset> #include <set> #include <queue> #include < stack> #include <map> #include <cstdlib> #include <cmath> #define LL long #define PB push_back #de Fine PA pair<int,int> #define CLR (a,b) memset (A,b,sizeof (a)) #define Lson lr<<1,l,mid #define Rson lr<< 1|1,mid+1,r #define BUG (x) printf ("%d++++++++++++++++++++%d\n", x,x) #define Key_value ch[ch[root][1]][0] #pragma
Comment (linker, "/stack:102400000000,102400000000") const LL MOD = 1000000007;
const int N = 100+15;
const int MAXN = 1E5+15;
const int letter = 130;
Const LL INF = 1e7;
Const double Pi=acos (-1.0);
const double EPS=1E-10;
using namespace Std;
    inline int read () {int X=0,f=1;char ch=getchar (); while (ch< ' 0 ' | |
    Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();} while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; ch=geTCHAR ();}
return x*f;
int m;
LL n,b[25],dp[25][25][25],bin[25];
    Dp[i][j][k] represents the maximum value of k multiplication in the i-j interval int main () {int TC;
    scanf ("%d", &AMP;TC);
    Bin[0]=1;
        while (tc--) {cin>>n>>m;
        m--;
        CLR (dp,0);
        int k=0;
        while (n) {b[++k]=n%10,n/=10;}
        Reverse (b+1,b+k+1);
        for (int i=1;i<=k;i++) dp[i][i][0]=b[i];
                for (int len=2;len<=k;len++) {for (int i=1;i<=k;i++) {int j=i+len-1;
                if (j>k) break;
                DP[I][J][0]=DP[I][J-1][0]*10+DP[J][J][0]; for (int x=1;x<=min (j-i,m); x + +) {for (int y=i;y<j;y++) {for (int z=0;z<=
                        Min (x-1,min (m,y-i)); z++) {Dp[i][j][x]=max (dp[i][j][x],dp[i][y][z]*dp[y+1][j][x-z-1]); }}}} cout<<dp[1][k][m]<<
    Endl
return 0;
 }