Four Operations
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0
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.
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
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
1 12345
Sample Output
Case #1: 1
为了满足条件乘号左右肯定是只有一个数,只要遍历除号的位置,定下了除号,那么减号,乘号,除号都已经确定,加号只可能在剩下的可放入的地方的最前或最后,取这些可能中的最大值即可。
#include<iostream>
#include<cmath>
#include<string.h>
using namespace std;
char num[33];
long long nn[33];
long long len;
long long _max;
long long run(long long p1,long long p2)
{
long long anss=0;
long long res=1;
for(int i=p2-1;i>=p1;i--)
{
anss+=nn[i]*res;
res*=10;
}
return anss;
}
void cal(long long a,long long b,long long c,long long d)
{
long long div=run(a,len);
long long chen=run(b,a);
long long jian=run(c,b);
long long jia=run(d,c);
long long x=run(0,d);
long long aa=x+jia-jian*chen/div;
//cout<<x<<" "<<jia<<' '<<jian<<' '<<chen<<' '<<div<<endl;
if(aa>_max)
{
_max=aa;
}
}
int main()
{
int cas;
scanf("%d",&cas);
long long cc=0;
while(cas--)
{
_max=-100000;
scanf("%s",&num);
len=strlen(num);
for(int i=0;i<len;i++)
{
nn[i]=num[i]-'0';
}
for(int i=len-1;i>=4;i--)
{
cal(i,i-1,i-2,1);
cal(i,i-1,i-2,i-3);
}
printf("Case #%lld: %lld\n",++cc,_max);
}
}