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2016年中国大学生程序设计竞赛（杭州）1006 Four Operations

2024-08-18 03:48:04   212人阅读

Four Operations

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38    Accepted Submission(s): 18


Problem Description
Little Ruins is a studious boy, recently he learned the four operations!

Now he want to use four operations to generate a number, he takes a string which only contains digits ‘1‘ - ‘9‘, and split it into 5 intervals and add the four operations ‘+‘, ‘-‘, ‘*‘ and ‘/‘ in order, then calculate the result(/ used as integer division).

Now please help him to get the largest result.
 

 

Input
First line contains an integer T, which indicates the number of test cases.

Every test contains one line with a string only contains digits ‘1‘-‘9‘.

Limits
1≤T≤105
5≤length of string≤20
 

 

Output
For every test case, you should output ‘Case #x: y‘, where x indicates the case number and counts from 1 and y is the result.
 

 

Sample Input
112345
 

 

Sample Output
Case #1: 1
 

 

Source
2016年中国大学生程序设计竞赛（杭州）
 

 

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/*枚举减号，刚开始天真的以为，除数最多是两位......卡死。*/#include<bits/stdc++.h>#define ll long long#define INF 0x3fffffffffffffff#define N 22using namespace std;string op;ll right(string s){    ll e=0;    int n=s.size();    for(int i=2;i<n;i++)        e=e*10+s[i]-‘0‘;    //cout<<c<<" "<<d<<" "<<e<<endl;    ll ans=(s[0]-‘0‘)*(s[1]-‘0‘)/e;    return ans;}ll left(string s)//减号左边的部分{    ll cur1=0,cur2=0;    int n=s.size();    for(int i=1;i<n;i++)        cur1=cur1*10+(s[i]-‘0‘);    cur1+=s[0]-‘0‘;    for(int i=0;i<n-1;i++)        cur2=cur2*10+(s[i]-‘0‘);    cur2+=s[n-1]-‘0‘;    //cout<<"max(cur1,cur2)="<<max(cur1,cur2)<<" ";    return max(cur1,cur2);}ll solve(string s){    ///枚举减号    ll cur=-INF,s1,s2;    int n=s.size();    for(int i=2;i<=n-3;i++)    {        s1=left(s.substr(0,i));        s2=right(s.substr(i,n-i));        //cout<<s1<<" "<<s2<<endl;        cur=max(cur,s1-s2);    }    printf("%lld\n",cur);}int t;int main(){    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);    scanf("%d",&t);    for(int Case=1;Case<=t;Case++)    {        printf("Case #%d: ",Case);        cin>>op;        solve(op);    }    return 0;}

 

2016年中国大学生程序设计竞赛（杭州）1006 Four Operations


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