九度oj--1002和1045

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题目1002：Grading

题目描述：
Grading hundreds of thousands of Graduate Entrance Exams is a hard work. It is even harder to design a process to make the results as fair as possible. One way is to assign each exam problem to 3 independent experts. If they do not agree to each other, a judge is invited to make the final decision. Now you are asked to write a program to help this process.
For each problem, there is a full-mark P and a tolerance T(

import java.util.*;public class Main {    public static void main(String args[]) {        Scanner cin = new Scanner(System.in);        int p,t,g1,g2,g3,gj,max;        double grade;        while(cin.hasNext()){           p = cin.nextInt();           t = cin.nextInt();           g1 = cin.nextInt();           g2 = cin.nextInt();           g3 = cin.nextInt();           gj = cin.nextInt();           grade = 0.0;           if(Math.abs(g1 - g2) <= t)                grade = (double)(g1 + g2)/2;           else if(Math.abs(g1-g3) <= t && Math.abs(g2-g3) <= t){                max = g1;                if(g2 > max)                    max = g2;                if(g3 > max)                    max = g3;                grade = max;                }            else if(Math.abs(g1 - g3) <= t)                grade = (double)(g1 + g3)/2;            else if(Math.abs(g2 - g3) <= t)                grade = (double)(g2 + g3)/2;            else                grade = gj;            System.out.printf("%.1f\n", grade);        }    }}

被这个题折磨了好久，一直都是Wrong Answer还找不到错误，最后发现又是不仔细，只注意了在最后一种情况下求平均数时转换double，没有注意到第一种情况也求了平均数，忘记了转换类型而一直出错。

题目1045：百鸡问题

题目描述：
用小于等于n元去买100只鸡，大鸡5元/只，小鸡3元/只,还有1/3元每只的一种小鸡，分别记为x只,y只,z只。编程求解x,y,z所有可能解。
输入：
测试数据有多组，输入n。
输出：
对于每组输入,请输出x,y,z所有可行解，按照x，y，z依次增大的顺序输出。
样例输入：
40
样例输出：
x=0,y=0,z=100
x=0,y=1,z=99
x=0,y=2,z=98
x=1,y=0,z=99

import java.util.*;public class Main {    public static void main(String args[]) {        Scanner cin = new Scanner(System.in);        int i,j,k,n,sum;        //double sum;        while(cin.hasNext()){            n = cin.nextInt();            for(i = 0 ; i <= n/5 ; i++){                for(j = 0 ; j <= n/3; j++){                    sum = n - 5*i - 3*j;                    k = sum * 3;                    if( (i + j + k) >= 100){                        k = 100 - i - j;                        System.out.printf("x=" + i + ",y=" + j + ",z=" + k + "\n");                        }                }            }        }    }}

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