倒水(贪心吧)

// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1041
// Memory Limit: 65536 MB
// Time Limit: 2000 ms
// 2022-03-03 20:18:54
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
const int N = 1e5 + 10;

double t[N], c[N];
int n;
void solve() {
	int　n;
    cin >> n;
    double T = 0, V = 0;
    cin >> T >> V;
    double v1 = T * V, c1 = V;
    double tmin = 1e9, tmax = 0;
    for (int i = 1; i<= n; i ++) {
        cin >> t[i] >> c[i];
        v1 += t[i] * c[i];
        c1 += c[i];
        tmin = min (tmin, t[i]);
        tmax = max (tmax, t[i]);
    }
    double a = v1 / c1;
    
    if (a >= tmax) {
        puts("Possible");
        printf("%.4f\n", a);
    }
    else if (a <= tmin) {
        puts("Possible");
        printf("%.4f\n", tmin);
    }
    else puts("Impossible");
}
int main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

开始陷入了对添加多少水具体的思考。但是太复杂了。 实际这题通过求平均温度。然后判断与水杯的的最高和最低温度的关系 如果<=min，那么说明水缸的温度比杯子中的任何一个都低，所以可以调节，反之类似