贪心 1013

Accounting for Computer Machinists (ACM) has sufferred from the Y2K bug and lost some vital data for preparing annual report for MS Inc.
All what they remember is that MS Inc. posted a surplus or a deficit each month of 1999 and each month when MS Inc. posted surplus, the amount of surplus was s and each month when MS Inc. posted deficit, the deficit was d. They do not remember which or how many months posted surplus or deficit. MS Inc., unlike other companies, posts their earnings for each consecutive 5 months during a year. ACM knows that each of these 8 postings reported a deficit but they do not know how much. The chief accountant is almost sure that MS Inc. was about to post surplus for the entire year of 1999. Almost but not quite.

Write a program, which decides whether MS Inc. suffered a deficit during 1999, or if a surplus for 1999 was possible, what is the maximum amount of surplus that they can post.

 

Input

Input is a sequence of lines, each containing two positive integers s and d.

 

Output

For each line of input, output one line containing either a single integer giving the amount of surplus for the entire year, or output Deficit if it is impossible.

 

Sample Input

59 237375 743200000 8496942500000 8000000

 

Sample Output

11628300612
Deficit
这个题 读题最麻烦
月盈利问题
16 27 38 49 510 611 712
五个月亏损
这就要计算
在每个月亏损的条件下
最大年盈利
f表示亏损 亏损要在最大的重叠月份 5月 10月
ttttf剩下循环
tttff
ttfff
tffff5个一循环这是4种情况 前五个月的
就是这样
然后判断
最后小于0返会deficit
#include<iostream>#include<string.h>#include<set>#include<stdio.h>#include<vector>#include<algorithm>#include<numeric>#include<math.h>#include<string.h>#include<sstream>#include<stdio.h>#include<string>#include<cstdlib>#include<algorithm>#include<iostream>#include<map>#include<queue>#include<string>using namespace std;int main(){    int n,p;    int sum;    while(cin>>n)    {        cin>>p;        if( p>n*4)        sum=10*n-2*p;        else if(2*p>3*n)              sum=8*n-4*p;        else if (3*p>2*n)              sum=6*n-6*p;        else if (4*p>n)              sum=3*n-9*p;              else              sum=-12*p;    if(sum<=0)    cout<<"Deficit"<<endl;    else    cout<<sum<<endl;    }    return 0;}