听了ZWK大大的思路，就立马1A了
思路是这样的：
算最小GPA的时候，首先每个科目分配到69分（不足的话直接输出GPA 2），然后FOR循环下来使REMAIN POINT减少，每个科目的上限加到100即可
算最大GPA的时候，首先每个科目分配到60分,然后FOR循环下来使REMAIN POINT减少，每个科目的上限加到85即可，如果还有REMAIN POINT，就FOR循环下来加到100上限即可
不会DP 阿 QAQ
过段时间得好好看DP了 = =
于是默默的把这题标记为《水题集》
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define ll long long
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x))) #define NN 200011 using namespace std; const ll INF = 0x3f3f3f3f;
const ll MAXN = ; int main(){
int i, j, n, m, k;
int numCase;
int score, sum;
int a[];
scanf("%d",&numCase);
while(numCase--){
double Gmax = 0.0;
double Gmin = 0.0;
scanf("%d%d",&score,&n);
sum = score * n; if(score < ) Gmax = ;
else{
for(i = ; i <= n; ++i){
a[i] = ;
}
int remain = sum - n * ;
while(remain != ){
for(i = ; i <= n; ++i){
if(a[i] <= ){
int delta = - a[i];
if(remain > delta){
a[i] += delta;
remain -= delta;
} else{
a[i] += remain;
remain = ;
break;
} }
}
for(i = ; i <= n; ++i){
if(a[i] <= ){
int delta = - a[i];
if(remain > delta){
a[i] += delta;
remain -= delta;
} else{
a[i] += remain;
remain = ;
break;
} }
}
}
for(i = ; i <= n; ++i){
if(a[i] >= ) Gmax += 4.0;
else if(a[i] >= ) Gmax += 3.5;
else if(a[i] >= ) Gmax += 3.0;
else if(a[i] >= ) Gmax += 2.5;
else if(a[i] >= ) Gmax += 2.0;
}
Gmax /= n;
} if(score < ) Gmin = ;
else{
for(i = ; i <= n; ++i){
a[i] = ;
}
int remain = sum - n * ;
while(remain != ){
for(i = ; i <= n; ++i){
if(a[i] <= ){
int delta = - a[i];
if(remain > delta){
a[i] += delta;
remain -= delta;
} else{
a[i] += remain;
remain = ;
break;
} }
}
}
for(i = ; i <= n; ++i){
if(a[i] >= ) Gmin += 4.0;
else if(a[i] >= ) Gmin += 3.5;
else if(a[i] >= ) Gmin += 3.0;
else if(a[i] >= ) Gmin += 2.5;
else if(a[i] >= ) Gmin += 2.0;
}
Gmin /= n;
}
printf("%.4f %.4f\n",Gmin,Gmax);
}
return ;
}