Number Theory-division-returns the maximum multiple of 3 in an integer set.

 

Mean: 

Description: An array containing non-negative integers (N in length) is used to find the maximum multiple of 3 consisting of these numbers. If not, impossible is output.

Analyze:

The first thing that comes to mind is direct violence. This is the dumbest method. If there is a large amount of data, it must be TLE.

Directly use brute force to generate all the combinations, which are 2 ^ N. Compare and judge each number. It takes O (n) time, because N may be relatively large, comparison of each bit is required.
The total time complexity is O (n * 2 ^ N ).

 

So what should we do?

First, let's learn a few mathematical knowledge:

1) if the remainder of a number n to m is A1, and B is the sum of each digit of N, there are: N % m = B % m = A1.

For example, for the sum of 151 and all of them 7, the remainder of 3 is 1.

2) If a number is a multiple of 3, the sum of the numbers is a multiple of 3.
For example, let's consider 8760, which is a multiple of 3, because the total number is 8 + 7 + 6 + 0 = 21, which is a multiple of 3.

Deduction: If a number is a multiple of 3, the number is arranged as a new number, and the number is still a multiple of 3.

 

Time Complexity:O (nlongn)

 

Source code:

Brute force code:

// Memory   Time// 1347K     0MS// by : Snarl_jsb// 2014-09-10-09.28#include<algorithm>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<vector>#include<queue>#include<stack>#include<map>#include<string>#include<climits>#include<cmath>#define N 100#define MAXN 100#define LL long longusing namespace std;long long n,summ;long long a[N];bool cmp(LL a,LL b){    return a>b;}LL ans[N];LL idx;LL Max;void work(int* b,int n){LL sum1=0;for(int i=0;i<n;i++)    {        sum1+=a[b[i]];    }    if(sum1%3==0&&sum1>Max)    {        Max=sum1;        idx=n;        for(int i=0;i<n;i++)        {            ans[i]=a[b[i]];        }    }}void _gen_comb(int* a,int s,int e,int m,int& cnt,int* temp){int i;if (!m)work(temp,cnt);elsefor (i=s;i<=e-m+1;i++){temp[cnt++]=a[i];_gen_comb(a,i+1,e,m-1,cnt,temp);cnt--;}}void gen_comb(int n,int m){int a[MAXN],temp[MAXN],cnt=0,i;for (i=0;i<n;i++)a[i]=i+1;_gen_comb(a,0,n-1,m,cnt,temp);}int main(){//    freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);//    freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);    while(~scanf("%I64d",&n))    {        LL sum=0;        for(int i=1;i<=n;++i)        {            scanf("%I64d",a+i);            sum+=a[i];        }        summ=sum;//        cout<<sum<<endl;        sort(a+1,a+1+n,cmp);        if(!(sum%3))        {            for(int i=1;i<=n;i++)            {                printf("%I64d",a[i]);            }            puts("");            continue;        }        bool flag=0;        for(int i=n;i>1;--i)        {            sum-=a[n];            if(!(sum%3))            {                for(int j=1;j<i;j++)                    printf("%I64d",a[j]);                puts("");                flag=1;                break;            }        }        if(flag) continue;        Max=LONG_LONG_MIN;        for(int i=n;i>=1;--i)        {            gen_comb(n,i);        }        if(sizeof(ans)==0)        {            puts("impossible\n");            continue;        }        sort(ans,ans+idx,cmp);        for(int i=0;i<idx;++i)        {            printf("%I64d",ans[i]);        }        puts("");    }    return 0;}

　　

Mathematical Methods:

// Memory   Time// 1347K     0MS// by : Snarl_jsb// 2014-09-10-11.37#include<algorithm>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<vector>#include<queue>#include<stack>#include<map>#include<string>#include<climits>#include<cmath>#define N 10000#define LL long longusing namespace std;LL n;LL a[N];LL ans[N];LL n1,n2,n3;LL q1[N],q2[N],q3[N];bool cmp(LL a,LL b){    return a>b;}int main(){//    freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);//    freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);    while(scanf("%d",&n)!=EOF)    {        n1=n2=n3=0;        LL sum=0;        for(LL i=1;i<=n;i++)        {            scanf("%d",&a[i]);            sum+=a[i];            if(a[i]%3==0)            {                q1[++n1]=a[i];            }            else if(a[i]%3==1)                q2[++n2]=a[i];            else                q3[++n3]=a[i];        }        if(sum%3==0)        {            sort(a+1,a+1+n);            for(int i=n;i>=1;--i)            {                printf("%d",a[i]);            }            puts("");            continue;        }        sort(q1+1,q1+1+n1,cmp);        sort(q2+1,q2+1+n2,cmp);        sort(q3+1,q3+1+n3,cmp);        if(sum%3==1)        {            if(n2>=1)            {                n2--;            }            else if(n3>=2)            {                n3-=2;            }            else            {                puts("impossible\n");                continue;            }        }        else if(sum%3==2)        {            if(n1>=2)            {                n1-=2;            }            else if(n2>=1)            {                n2--;            }            else            {                puts("impossible\n");                continue;            }        }        LL idx=0;        for(int i=1;i<=n1;++i)        {            ans[++idx]=q1[i];        }        for(int i=1;i<=n2;++i)        {            ans[++idx]=q2[i];        }        for(int i=1;i<=n3;++i)        {            ans[++idx]=q3[i];        }        sort(ans+1,ans+1+idx,cmp);        for(int i=1;i<=idx;i++)        {            printf("%I64d",ans[i]);        }        puts("");    }    return 0;}

　　

Number Theory-division-returns the maximum multiple of 3 in an integer set.