2015 ACM/ICPC Asia Regional Shenyang Online

1012 Largest Point

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4804    Accepted Submission(s): 881

Problem Description

Given the sequence A with n integers t1,t2,⋯,tn. Given the integral coefficients a and b. The fact that select two elements ti and tj of A and i≠j to maximize the value of at2i+btj, becomes the largest point.

 

Input

An positive integer T, indicating there are T test cases.
For each test case, the first line contains three integers corresponding to n (2≤n≤5×106), a (0≤|a|≤106) and b (0≤|b|≤106). The second line contains n integers t1,t2,⋯,tn where 0≤|ti|≤106 for 1≤i≤n.

The sum of n for all cases would not be larger than 5×106.

 

Output

The output contains exactly T lines.
For each test case, you should output the maximum value of at2i+btj.

 

Sample Input

2

3 2 1
1 2 3

5 -1 0
-3 -3 0 3 3

 

Sample Output

Case #1: 20
Case #2: 0

 

各位童鞋千万注意，输入ti时也要用long long，比赛时wa 20+

#include <bits/stdc++.h>
#define LL long long
using namespace std;
LL t[5000010];
typedef struct node
{
    LL data;
    int id;
    bool operator <(const node &x)const
    {
        return data > x.data;
    }
} FF;
FF L[5000010],R[5001000];
int main()
{
    int T,n,a,b;
    int cas = 0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&a,&b);
        for(int i=0; i<n; i++)
        {
            scanf("%lld",&t[i]);
            L[i].data = t[i] * t[i] * a;
            R[i].data = t[i] * b;
            L[i].id = i;
            R[i].id = i;
        }
        sort(L,L+n);
        sort(R,R+n);
        LL ans;
        if(L[0].data > R[0].data)
        {
            ans = L[0].data;
            if(L[0].id != R[0].id)
                ans += R[0].data;
            else
                ans += R[1].data;
        }
        else
        {
            ans = R[0].data;
            if(L[0].id != R[0].id)
                ans += L[0].data;
            else
                ans += L[1].data;
        }

        printf("Case #%d: %lld\n",++cas,ans);
    }
    return 0;
}

1006 Fang Fang

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4656    Accepted Submission(s): 986

Problem Description

Fang Fang says she wants to be remembered.
I promise her. We define the sequence F of strings.
F0 = ‘‘f",
F1 = ‘‘ff",
F2 = ‘‘cff",
Fn = Fn−1 + ‘‘f", for n > 2
Write down a serenade as a lowercase string S in a circle, in a loop that never ends.
Spell the serenade using the minimum number of strings in F, or nothing could be done but put her away in cold wilderness.

 

Input

An positive integer T, indicating there are T test cases.
Following are T lines, each line contains an string S as introduced above.
The total length of strings for all test cases would not be larger than 106.

 

Output

The output contains exactly T lines.
For each test case, if one can not spell the serenade by using the strings in F, output −1. Otherwise, output the minimum number of strings in F to split S according to aforementioned rules. Repetitive strings should be counted repeatedly.

 

Sample Input

8
ffcfffcffcff
cffcfff
cffcff
cffcf
ffffcffcfff
cffcfffcffffcfffff
cff
cffc

 

Sample Output

Case #1: 3
Case #2: 2
Case #3: 2
Case #4: -1
Case #5: 2
Case #6: 4
Case #7: 1
Case #8: -1
Hint
Shift the string in the first test case, we will get the string "cffffcfffcff"
and it can be split into "cffff", "cfff" and "cff".

#include <bits/stdc++.h>
using namespace std;
char s[1000000+10];
int pos[1000000+10];
int e;
int main()
{
    int tg;
    scanf("%d%*c", &tg);
    int len, i, j;
    int cnt=1;
    while(tg--)
    {
        gets(s);
        len=strlen(s);
        if(len==0)
        {
            printf("Case #%d: 0\n",cnt++);
            continue;
        }
        bool ss=true;
        for(i=0; i<len; i++)
        {
            if(s[i]!='c' && s[i]!='f')
            {
                ss=false;
                break;
            }
        }
        if(ss==false)
        {
            printf("Case #%d: -1\n", cnt++);
            continue;
        }
        bool ok=false;
        e=0;
        for(i=0; i<len; i++)
        {
            if(s[i]=='c')
            {
                ok=true;
                pos[e++]=i;
            }
        }
        if(ok==false)
        {
            if(len%2==0)
                printf("Case #%d: %d\n", cnt++, len/2);
            else
                printf("Case #%d: %d\n", cnt++, len/2+1);
        }
        else
        {
            if(e*2 > (len-e) )
                printf("Case #%d: -1\n", cnt++);
            else
            {
                pos[e]=len+pos[0];
                bool z=true;
                for(i=0; i<e; i++)
                {
                    if(pos[i+1]-pos[i]<=2)
                    {
                        z=false;
                        break;
                    }
                }
                if(z==false)
                    printf("Case #%d: -1\n", cnt++);
                else
                    printf("Case #%d: %d\n", cnt++, e);
            }
        }
    }
    return 0;
}