Problem Description
Xueba: Using the 4-Point Scale, my GPA is 4.0.
In fact, the AVERAGE SCORE of Xueba is calculated by the following formula:
AVERAGE SCORE = ∑(Wi * SCOREi) / ∑(Wi) 1<=i<=N
where SCOREi represents the scores of the ith course and Wi represents the credit of the corresponding course.
To simplify the problem, we assume that the credit of each course is 1. In this way, the AVERAGE SCORE is ∑(SCOREi) / N. In addition, SCOREi are all integers between 60 and 100, and we guarantee that ∑(SCOREi) can be divided by N.
In SYSU, the university usually uses the AVERAGE SCORE as the standard to represent the students’ level. However, when the students want to study further in foreign countries, other universities will use the 4-Point Scale to represent the students’ level. There
are 2 ways of transforming each score to 4-Point Scale. Here is one of them.
The student’s average GPA in the 4-Point Scale is calculated as follows:
GPA = ∑(GPAi) / N
So given one student’s AVERAGE SCORE and the number of the courses, there are many different possible values in the 4-Point Scale. Please calculate the minimum and maximum value of the GPA in the 4-Point Scale.
Input
The input begins with a line containing an integer T (1 < T < 500), which denotes the number of test cases. The next T lines each contain two integers AVGSCORE, N (60 <= AVGSCORE <= 100, 1 <= N <= 10).
Output
For each test case, you should display the minimum and maximum value of the GPA in the 4-Point Scale in one line, accurate up to 4 decimal places. There is a space between two values.
Sample Input
4 75 1 75 2 75 3 75 10
Sample Output
3.0000 3.0000 2.7500 3.0000 2.6667 3.1667 2.4000 3.2000 Hint In the third case, there are many possible ways to calculate the minimum value of the GPA in the 4-Point Scale. For example, Scores 78 74 73 GPA = (3.0 + 2.5 + 2.5) / 3 = 2.6667 Scores 79 78 68 GPA = (3.0 + 3.0 + 2.0) / 3 = 2.6667 Scores 84 74 67 GPA = (3.5 + 2.5 + 2.0) / 3 = 2.6667 Scores 100 64 61 GPA = (4.0 + 2.0 + 2.0) / 3 = 2.6667
贪心加暴力，贪心就是 求最大值时只要取分数的下界，最小值只取上界，然后直接暴力枚举每个边界的个数，判断是否合法，更新答案。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <set>
#include <stack>
#include <cctype>
#include <algorithm>
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef long long LL;
const int mod = 99999997;
const int MAX = 0x3f3f3f3f;
const int maxn = 100005;
const int N = 100005;
int t, n, av, tot;
bool check1(int i1, int i2, int i3, int i4, int i5) {
if(i5 < 0) return false;
int ans = i1*85 + i2*80 + i3*75 + i4*70 + i5*60;
if(ans <= tot) return true;
return false;
}
bool check2(int i1, int i2, int i3, int i4, int i5) {
if(i5 < 0) return false;
int ans = i1*100 + i2*84 + i3*79 + i4*74 + i5*69;
if(ans >= tot) return true;
return false;
}
double F1(double i1, double i2, double i3, double i4, double i5) {
return i1*4 + i2*3.5 + i3*3 + i4*2.5 + i5*2;
}
int main()
{
cin >> t;
while(t--) {
cin >> av >> n;
tot = av*n;
double ans1 = 10000000000000;
for(int i1 = 0; i1 <= n; i1++)
for(int i2 = 0; i1 + i2 <= n; i2++)
for(int i3 = 0; i1 + i2 + i3 <= n; i3++)
for(int i4 = 0; i1 + i2 + i3 + i4 <= n; i4++) {
int i5 = n - i1 - i2 - i3 - i4;
if(check2(i1, i2, i3, i4, i5)) {
double sum = F1(i1, i2, i3, i4, i5);
ans1 = min(ans1, sum);
}
}
printf("%.4lf ", ans1/n);
double ans = 0;
for(int i1 = 0; i1 <= n; i1++)
for(int i2 = 0; i1 + i2 <= n; i2++)
for(int i3 = 0; i1 + i2 + i3 <= n; i3++)
for(int i4 = 0; i1 + i2 + i3 + i4 <= n; i4++) {
int i5 = n - i1 - i2 - i3 - i4;
if(check1(i1, i2, i3, i4, i5)) {
double sum = F1(i1, i2, i3, i4, i5);
ans = max(ans, sum);
}
}
printf("%.4lf\n", ans/n);
}
return 0;
}
HDU 4968 Improving the GPA 多校第九场1009