c++控制小数精度和四舍五入

以防自己忘记，作此笔记，先说结论：

#include <iostream>
#include <iomanip>
using namespace std;
int main() {
	double k1 = 1534.453;
	double k2 = 1534.458;
	cout << fixed << setprecision(2) << k1 << endl;
	//cout << 表明自己输出的小数要固定字数 << 保留几位，不够的也会补0 << 浮点数 << endl;
	cout << fixed << setprecision(2) << k2 << endl;
	return 0;
}

这个输出结果分别是1534.45，1534.46。结果会自动四舍五入。

做题错误来源于：P11046 [蓝桥杯 2024 省 Java B] 星际旅行 - 洛谷

当时不知道这些，直接用字符串模拟，结果只有部分分数，部分结果是错误的：

#include <iostream>
#include <vector>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
using namespace std;
#define endl '\n'
#define int long long
const int INF = 0x3f3f3f3f;

int n, m, q;
vector<int> vis;
vector<int> g[1010];

int bfs(int p, int times) {
    vis.assign(n + 4, 0);
    int cnt = 0;
    queue<int> q;
    q.push(p);
    vis[p] = 1;
    cnt++;
    while (q.size() && times) {
        int p = q.size();
        while (p--) {
            int x = q.front();
            q.pop();
            for (int ch : g[x]) {
                if (vis[ch] == 0) {
                    q.push(ch);
                    vis[ch] = 1;
                    cnt++;
                }
            }
        }
        times--;
    }
    return cnt;
}

void solve() {
    cin >> n >> m >> q;
    int ans = 0;
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    for (int i = 0; i < q; i++) {
        int a, times;
        cin >> a >> times;
        ans += bfs(a, times);
    }
    double t = (double)ans * 1.0 / q;
    string s = to_string(t);
    int pos = s.find('.');
    pos += 3;
    int flag = 0;
    int check = s[pos] - '0';
    if (check >= 5) {
        flag = 1;
    }
    if (flag == 1) {
        t += 0.01;
    }
    printf("%.2f\n", t);
}

signed main()
{
    int t;
    //cin >> t;
    t = 1;
    while (t--) {
        solve();
    }
    return 0;
}

正解还得用笔记内容：

#include <iostream>
#include <vector>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <iomanip>
using namespace std;
#define endl '\n'
#define int long long
const int INF = 0x3f3f3f3f;

int n, m, q;
vector<int> vis;
vector<int> g[1010];

int bfs(int p, int times) {
    vis.assign(n + 4, 0);
    int cnt = 0;
    queue<int> q;
    q.push(p);
    vis[p] = 1;
    cnt++;
    while (q.size() && times) {
        int p = q.size();
        while (p--) {
            int x = q.front();
            q.pop();
            for (int ch : g[x]) {
                if (vis[ch] == 0) {
                    q.push(ch);
                    vis[ch] = 1;
                    cnt++;
                }
            }
        }
        times--;
    }
    return cnt;
}

void solve() {
    cin >> n >> m >> q;
    int ans = 0;
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    for (int i = 0; i < q; i++) {
        int a, times;
        cin >> a >> times;
        ans += bfs(a, times);
    }
    double expectation = (double)ans / q;
    cout << fixed << setprecision(2) << expectation << endl;
}

signed main()
{
    int t;
    //cin >> t;
    t = 1;
    while (t--) {
        solve();
    }
    return 0;
}