leetcode动态规划切钢筋

钢筋的价值和长度有关，以下为初始数据

    v[0] = 0;
    v[1] = 1;
    v[2] = 5;
    v[3] = 8;
    v[4] = 9;//不是最大值，注意
    v[5] = 10;
    v[6] = 20;
    v[7] = 20;
    v[8] = 20;
    v[9] = 20;
    v[10] = 20;

思路

1. 动态规划，只管最后一刀切的位置
2. 需要将n之前的长度的价值最大值求取出来，进行累计

#include<stdio.h>
#include <string.h>
#define MAXLEN 1000
#define MAX(a,b) ((a) >(b)? a:b)
void  MaxValueC(int n,int *v,int k){
    if(n==0 || n==1||n==2|| n==3) {
        return  ;
    
    } else {
        for(int j = 0;j<=n;j++)
        for (int i = 0; i<=j&&i<=10; i++) {
            v[j] = MAX(v[i]+v[j-i],v[j]);
        }
    } 
}

int MaxValue(int n) {
    int v[MAXLEN];
    memset(v, 0,sizeof(v));
    int k = 0;
    v[0] = 0;
    v[1] = 1;
    v[2] = 5;
    v[3] = 8;
    v[4] = 9;
    v[5] = 10;
    v[6] = 20;
    v[7] = 20;
    v[8] = 20;
    v[9] = 20;
    v[10] = 20;

    MaxValueC(n,v,k);
    for (int i=0; i<=n; i++) {
        printf("%d ",v[i]);
    }
    printf("\n");
    return v[n];
}


int main () {
    int n = 30;
    int re = MaxValue(n);
    printf("re = %d\n",re);

}

标签：10,20,MaxValueC,int,钢筋,re,printf,动态,leetcode
来源： https://blog.csdn.net/weixin_41672404/article/details/120219092