POJ 2586：Y2K Accounting Bug：贪心法

Y2K Accounting Bug

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9519 Accepted: 4753

Description

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

11628300612Deficit

Source

Waterloo local 2000.01.29

理解题意以后，很容易想到，贪心法前5个月满足最小的亏损，并且盈利在前面，亏损在后面。后面的7个月只要保证前面的5个月是最小亏损就可以。然后把12个月的亏损盈利加在一起，就得出盈利还是亏损。

#include<stdio.h>#include<math.h>__int64 s,d,sum,a[15];void chuli(){int i,j;int p,q;__int64 ss=0;p=i=0;q=4;for(j=0;j<5;j++){if(ss+s>=0){a[q--]=-1*d;ss-=d;}else{a[p++]=s;ss+=s;}}sum=ss;for(;j<12;i++,j++){ss-=a[i];if(ss+s>=0){a[j]=-1*d;ss-=d;}else{a[j]=s;ss+=s;}sum+=a[j];}}int main(){int i,j;while(scanf("%lld%lld",&s,&d)!=EOF){sum=0;chuli();if(sum>0)printf("%lld\n",sum);elseif(sum<0)printf("Deficit\n");}return 0;}