get

• 被精度坑死了（B，C).遇到精度問題看能不能轉(zhuǎn)化為整數(shù)，不行的話就用eps判斷

A. If at first you don’t succeed…

題意

有兩個(gè)聚會(huì)，A個(gè)人去第一個(gè)聚會(huì)，B個(gè)人去第二個(gè)聚會(huì)，C個(gè)人兩個(gè)都去了，一共有N個(gè)人，至少有一個(gè)人一個(gè)聚會(huì)也沒有去，問一共有幾個(gè)人沒有去聚會(huì)，如果無解輸出-1

AC

#include <bits/stdc++.h> #define N 200005 #define ll long long using namespace std; int main() {// freopen("in.txt", "r", stdin);ios::sync_with_stdio(false);int a, b, c, n;cin >> a >> b >> c >> n;if (a + b - c >= n || c >= n || a < c || b < c) cout << -1 << endl;else cout << n - (a + b - c) << endl;return 0; }

B. Getting an A

題意

給一個(gè)長(zhǎng)度為N序列，每個(gè)數(shù)最大為5。為改變幾個(gè)數(shù)可以讓平均數(shù)為5（四舍五入）
因?yàn)榫缺籬ack

// 轉(zhuǎn)化為整數(shù)處理 #include <bits/stdc++.h> #define N 200005 using namespace std; long long a[N], b[N]; vector<int> v; // pair<int,int> sum[N]; int main() {// freopen("in.txt", "r", stdin);ios::sync_with_stdio(false);int n;cin >> n;vector<int> v(n);for (int i = 0; i < n; i++) {cin >> v[i];}int sum = accumulate(v.begin(), v.end(), 0);int tar = (5 * n - n / 2);sort(v.begin(), v.end());if (sum >= tar) {cout << 0 << endl;return 0;}int ans = 0;for (auto it : v) {sum += (5 - it);ans++;if (sum >= tar) break;}cout << ans << endl;return 0; } #include <bits/stdc++.h> #define N 200005 using namespace std; long long a[N], b[N]; vector<int> v; // pair<int,int> sum[N]; int main() {// freopen("in.txt", "r", stdin);ios::sync_with_stdio(false);int n;cin >> n;vector<int> v(n);for (int i = 0; i < n; i++) {cin >> v[i];}sort(v.begin(), v.end());int sum = accumulate(v.begin(), v.end(), 0);if (sum * 1.0 / n >= 4.5) {cout << 0 << endl;return 0;}int ans = 0;for (int i = 0; i < n; i++) {sum += 5 - v[i];ans++;if (sum * 1.0 / n >= 4.5) break;// cout << i << " " << sum << endl;}cout << ans << endl;return 0; } // 精度錯(cuò)誤 #include <bits/stdc++.h> #define N 200005 using namespace std; long long a[N], b[N]; double eps = 1e-5; vector<int> v; // pair<int,int> sum[N]; int main() {// freopen("in.txt", "r", stdin);ios::sync_with_stdio(false);int n;cin >> n;vector<int> v(n);for (int i = 0; i < n; i++) {cin >> v[i];}sort(v.begin(), v.end());int sum = accumulate(v.begin(), v.end(), 0);double now = sum * 1.0 / n;if ( now >= 4.5) {cout << 0 << endl;return 0;}int ans = 0;for (int i = 0; i < n; i++) {// 含有小數(shù)的運(yùn)算，容易出現(xiàn)精度問題now += (5 - v[i]) * 1.0 / n;ans++;if (now >= 4.5 || fabs(now - 4.5) <= eps) break;// cout << i << " " << sum << endl;}cout << ans << endl;return 0; }

C. Candies

題意

一共有N個(gè)糖果，Vasya每次吃K個(gè)，Petya吃剩下的%10 (向下取整)，求在Vasya至少吃一半糖果的情況下最小的K

AC

// 二分 include <bits/stdc++.h> #define N 200005 #define ll long long using namespace std; bool solve(ll n, ll m) {ll sum_l = 0, sum_r = 0;while (n > 0) {if (n >= m)sum_l += m;elsesum_l += n;n -= m;if (n >= 10)sum_r += n / 10;n -= n / 10;}if (sum_l >= sum_r) return true;else return false; } int main() {// freopen("in.txt", "r", stdin);ios::sync_with_stdio(false);ll n;cin >> n;ll l = 1, r = n;while (l < r) {ll mid = (l + r) / 2;if (solve(n, mid))r = mid;elsel = mid + 1;}cout << l << endl;return 0; }

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