Inserting a plus (DP) dynamic plan into a string

Title: A string with a length of M inserts n plus signs for minimum and. For example, string str= "123456", n=2, Output 12+34+56 and 102, and Output 2 4, which is the plus position. Here is the idea:

The implementation process is mainly the idea of interval DP, wherein the state transfer equation is dp[i][j] = min (dp[k][j-1] + getnum (str.substr (k, i-k)), dp[i][j]);//where dp[i][j] means that the J plus sign is used in 1-i , where I use a struct is mainly to record the position of the plus.

#include <iostream> #include <ctime> #include <vector> #include <string> #include <algorithm > #include <sstream>using namespace std;//converts a string into a corresponding integer such as the Getnum value of the string "123" is 123int getnum (string str) {if ( Str.size () <= 0) return 0;int sum = 0;for (int i = 0; i < str.size (); i++) {sum = ten * sum + (Str[i]-' 0 ');} return sum;} Adding each number in the string, for example "123", is Getnum1 6int getnum1 (string str) {int sum = 0;for (int i = 0; i < str.size (); i++) {sum = sum + ( Str[i]-' 0 ');} return sum;} struct Node{int value;int k;}; int main () {string str = "1234567891"; int n = 3;int m = str.size (); VECTOR&LT;VECTOR&LT;NODE&GT;&GT;DP (M + 1, Vector<node > (n+1)); for (int i = 0, i < m+1; i++) {for (int j = 0; J < n+1; J + +) {dp[i][j].value = 9999999;DP[I][J].K = 0;}} vector<int>recordpos;//record the position of the plus sign dp[0][0].value = 0;int index = 0;for (int i = 0; I <=m; i++) {dp[i][0].value = Getnu M (Str.substr (0,i));} for (int j = 0; J <=n; J + +) {dp[0][j].value = 0;} for (int i = 1; i < m+1; i++) {for (iNT J = 1; j< n+1; J + +) {if (J >= i-1) Dp[i][j].value = getnum1 (str.substr (0, I)), else{for (int k = 1; k <=i; k++) {//dp[i][j] = min (dp[ K][j-1] + getnum (str.substr (k, i-k)), Dp[i][j]),//where dp[i][j] means that the 1-i is using the J Plus if (Dp[i][j].value>dp[k][j-1].value + get Num (Str.substr (k, i-k))) {Dp[i][j].value = Dp[k][j-1].value + Getnum (str.substr (k, i-k));//if (i==m) Recordpos.push_ba CK (k);DP [I][J].K = k;}}}} cout << dp[m][n].value << endl;cout << dp[m][n].k << endl;cout << dp[4][1].k << endl;i NT k = Dp[m][n].k;int y = n;while (k>0 && n>0) {recordpos.push_back (k); y--;m = K;k = DP[M][Y].K;} for (int i = 0; i < recordpos.size (); i++) {cout << recordpos[i] << "";} cout << Endl;cin.get (); return 0;}

Inserting a plus (DP) dynamic plan into a string