HDOJ 1024 Max Sum Plus

Idea: the dynamic planning method is used to solve the sum of the maximum m sub-segments. Because the data volume of this question is large, two-dimensional dp programming is not available and optimization is considered.
Using a temp [1000005] to store the largest sum of the I sub-segments of the first j elements, dp [j] = max (dp [J-1] + s [j], temp [J-1] + s [j]) represents the maximum child segment and

Note: scanf is widely used for input data.

Code:
# Include <iostream>
# Include <algorithm>
# Include <map>
# Include <cstdio>
# Include <string>
# Include <cstring>
Using namespace std;
Int s [1000005];
Int dp [1000005];
Int temp [1000005];
Void DP (int m, int n ){
Int I, j, k;
Int max;
Memset (dp, 0, sizeof (dp ));
Memset (temp, 0, sizeof (temp ));
For (I = 1; I <= m; I ++) {// use a for loop Traversal
Max =-999999999; // note that the value of max here must be an infinitely small value, which is initialized respectively when the maximum size of the I sub-segment is obtained.
For (j = I; j <= n; j ++ ){
Dp [j] = (dp [J-1] + s [j])> (temp [J-1] + s [j])? (Dp [J-1] + s [j]): (temp [J-1] + s [j]);
Temp [J-1] = max; // here temp [J-1] is used to store the element I sub-segment maximum of the previous J-1, note that it is not temp [j] Because max <dp [j] has not been determined;
If (max <dp [j])
Max = dp [j]; // update max
}
}
Cout <max <endl;
}
Int main (){
Int I, j;
Int n, m;
While (scanf ("% d", & m, & n )! = EOF ){
For (I = 1; I <= n; I ++)
Scanf ("% d", & s [I]);
DP (m, n );
}
Return 0;
}
Author: cgl1079743846