ZOJ 1110 Dick and Jane

Dick and Jane

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Time Limit: 2 Seconds       Memory Limit: 65536 KB

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Dick is 12 years old. When we say this, we mean that it is at least twelve and not yet thirteen years since Dick was born.

Dick and Jane have three pets: Spot the dog, Puff the Cat, and Yertle the Turtle. Spot was s years old when Puff was born; Puff was p years old when Yertle was born; Spot was y years old when Yertle was born. The sum of Spot's age, Puff's age, and Yertle's age equals the sum of Dick's age (d) and Jane's age (j). How old are Spot, Puff, and Yertle?

Each input line contains four non-negative integers: s, p, y, j. For each input line, print a line containing three integers: Spot's age, Puff's age, and Yertle's age. Ages are given in years, as described in the first paragraph.

Sample Input

5 5 10 9
5 5 10 10
5 5 11 10

Output for Sample Input

12 7 2
13 7 2
13 7 2

 题意：就是计算年龄。

 思路：以最后出生的yertle为基准，因为yertle最后出生，所以过了多少年就是多少岁，不存在进位的问题
按理想状态，设yertle为i岁，那么sp为i+y,pu为i+p,然后12+j-3*i-y-p就是我们要求的i的表达式
若这个式子被3整除，那么大家都正好，否则要么sp少1或者pu少1或者都少1.
若y-p==s那么说明sp没进位，那么再经相同的年数，sp比pu要先进位，所以此时sp++;
若y-p>s那么说明sp进位了，那么经过相同的年数，pu比sp先进位，所以pu++

代码：

#include


int main()
{
    int s,p,y,j,d;
    int sp,pu,ye;
    while(scanf("%d%d%d%d",&s,&p,&y,&j)!=EOF)
    {
        d=12+j-p-y;
        ye=d/3;
        pu=ye+p;
        sp=ye+y;
        if(d%3==1)
        {
            if(y-p>s)
                pu++;
            else
                sp++;
        }
        else if(d%3>1)
        {
            pu++;
            sp++;
        }
        printf("%d %d %d\n",sp,pu,ye);
    }
   return 0;
}