TOJ3976 Change 完全背包 || 找规律

After Shawn sees the following picture, he decides to give up his career in IT and turn to sell fruits. Since Shawn is a lazy guy, he doesn’t like to do any extra work. When he starts selling fruits, he finds that he always needs to take changes for customers. He wants you to write a program to give the least number of changes. Shawn is also a strange guy, because he takes changes by 6,5,3,1.

Input

The first line of input is an integer T which indicates the sum of test case. Each of the following T lines contains an integer n(1≤n≤10⁵), which is the total money Shawn need to give to the customer.

Output

For each test case, output "Case i: Result" in one line where i is the case number andResult is the least number of changes Shawn need to give customer.

Sample Input

341019

Sample output

Case 1: 2Case 2: 2Case 3: 4

Hint

Case 1: 4=3+1Case 2: 10=5+5Case 3: 19=6+6+6+1

第一反应是完全背包，无奈写炸TLE：

f[i][v]=max{f[i-1][v-k*c[i]]+k*w[i]，f[i][v]|0<=k*c[i]<=v}

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;int main(){    int T,V;    int w[4] = {6,5,3,1};    int dp[100005];    cin >> T;    for (int num = 1;num <= T;num++){        cin >> V;        for (int i = 0;i <= V;i++) dp[i] = i;        for (int i = 0;i < 4;i++){            for (int k = 0;k * w[i] <= V;k++){                for (int j = V;j >= k * w[i];j--){                    dp[j] = min(dp[j],dp[j - k * w[i]]+ k);                }            }        }        printf("Case %d: %d\n",num,dp[V]);    }    return 0;}

查题解发现可以直接套规律......

#include <stdio.h>int a[12]={1,2,1,2,1,1,2,2,2,2,2,2};int main(){int t,n,sum;scanf("%d",&t);for(int i=0;i<t;i++){scanf("%d",&n);sum=0;while(n>12){n-=6;sum++;}sum+=a[n-1];printf("Case %d: %d\n",i+1,sum); }}

后来仔细研读完全背包，发现应该优化为01背包.....看来本菜真的是忘完了（背包详情讲解看另一篇《背包九讲》）

procedure CompletePack(cost,weight)

   forv=cost..V

       f[v]=max{f[v],f[v-cost]+weight}

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;int main(){    int T,V;    int w[4] = {6,5,3,1};    int dp[100005];    cin >> T;    for (int num = 1;num <= T;num++){        cin >> V;        for (int i = 0;i <= V;i++) dp[i] = i;        for (int i = 0;i < 4;i++){            for (int j = w[i];j <= V;j++){                dp[j] = min(dp[j],dp[j-w[i]]+1);            }        }        printf("Case %d: %d\n",num,dp[V]);    }    return 0;}