Slingerspir
Lv.1
I only remember that I worked hard for 3 years.
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签名:Infinity
Lv.1
I only remember that I worked hard for 3 years.
签名:Infinity
在 我的电脑蓝屏啦！！！ 中回复
<html><head>
<title>PRANK: Windows 8 Update</title>
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background-size: cover;
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vertical-align:middle;
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text-align:center;
user-select:none;-moz-user-select:none;-webkit-user-select:none;-ms-user-select:none;
z-index:999; }
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position: relative;
padding-top: 100px;
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margin: auto;
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position: absolute;
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animation-name: orbit;
animation-duration: 5.5s;
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content: '';
position: absolute;
width: 4px;
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<a id="ref" data-translate="_win7configuringupdates">Configuring Windows Updates</a><br><a id="timer">2%</a> <span data-translate="_win7percent">complete.</span><br><span data-translate="_win7donotturnoff">Do not turn off your computer.</span>
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</body></html>2022-03-27T14:53:32 点赞:0
在 我会C++可以吗 中回复
2279. 数字变换
给定一个包含 5 个数字（0∼9）的字符串，例如 02943 ，请将 12345 变换到它。 你可以采取 3 种操作进行变换
交换相邻的两个数字。
将一个数字加 1 。如果加 1 后大于 9 ，则变为 0。
将一个数字加倍。如果加倍后大于 9 ,则将其变为加倍后的结果除以 10的余数。
最多只能用第 2 种操作 3 次，第 3 种操作 2次 求最少经过多少次操作可以完成变换。
){kind=link}
#include <iostream>
#include <string>
#include <queue>
#include <unordered_map>
using namespace std;
// 定义状态结构体
struct State {
string digits;
int steps;
State(string d, int s) : digits(d), steps(s) {}
};
int bfs() {
string target = "12345"; // 目标状态
unordered_map<string, int> stepsMap; // 保存每个状态所需的最小步数
queue<State> q;
q.push(State(target, 0));
stepsMap[target] = 0;
while (!q.empty()) {
State current = q.front();
q.pop();
if (current.digits == "12345") {
return current.steps; // 返回最小步数
}
for (int i = 0; i < 5; ++i) {
string next = current.digits;
// 操作1：交换相邻两个数字
swap(next[i], next[i+1]);
if (stepsMap.find(next) == stepsMap.end()) {
stepsMap[next] = current.steps + 1;
q.push(State(next, current.steps + 1));
}
// 操作2：将一个数字加1
next = current.digits;
next[i] = (next[i] - '0' + 1) % 10 + '0';
if (stepsMap.find(next) == stepsMap.end()) {
stepsMap[next] = current.steps + 1;
q.push(State(next, current.steps + 1));
}
// 操作3：将一个数字加倍并取余
next = current.digits;
next[i] = ((next[i] - '0') * 2) % 10 + '0';
if (stepsMap.find(next) == stepsMap.end()) {
stepsMap[next] = current.steps + 1;
q.push(State(next, current.steps + 1));
}
}
}
return -1; // 未找到解
}
int main() {
string input;
while (cin >> input) {
int result = bfs();
cout << result << endl;
}
return 0;
}
2024-05-06T21:38:59 点赞:0
在 我会C++可以吗 中回复
今天来改一下1元2次方程啊
学成归来
有问题求指出
#include <iostream>
#include <cmath>
using namespace std;
int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
void output_rational(double v) {
int p = round(v * 1000000);
int q = 1000000;
int d = gcd(p, q);
p /= d;
q /= d;
if (q == 1) {
cout << p << endl;
} else {
cout << p << "/" << q << endl;
}
}
int main() {
int T, M;
cin >> T >> M;
for (int i = 0; i < T; ++i) {
int a, b, c;
cin >> a >> b >> c;
int delta = b * b - 4 * a * c;
if (delta < 0) {
cout << "NO" << endl;
} else {
double x1 = (-b + sqrt(delta)) / (2.0 * a);
double x2 = (-b - sqrt(delta)) / (2.0 * a);
if (abs(x1) > abs(x2)) {
if (floor(x1) == x1) {
output_rational(x1);
} else {
int r = round(x1 * x1);
int q1 = floor(x1);
int q2 = round(1 / (x1 - q1));
int q3 = gcd(q2, r);
if (q1 != 0) {
output_rational(q1);
cout << "+";
}
if (q2 == 1) {
cout << "sqrt(" << r << ")" << endl;
} else if (q2 * q3 == 1) {
cout << q2 << "*sqrt(" << r << ")" << endl;
} else {
int c = q2 / q3;
int d = 1;
cout << c << "*sqrt(" << r << ")/" << d << endl;
}
}
} else {
if (floor(x2) == x2) {
output_rational(x2);
} else {
int r = round(x2 * x2);
int q1 = floor(x2);
int q2 = round(1 / (x2 - q1));
int q3 = gcd(q2, r);
if (q1 != 0) {
output_rational(q1);
cout << "+";
}
if (q2 == 1) {
cout << "sqrt(" << r << ")" << endl;
} else if (q2 * q3 == 1) {
cout << q2 << "*sqrt(" << r << ")" << endl;
} else {
int c = q2 / q3;
int d = 1;
cout << c << "*sqrt(" << r << ")/" << d << endl;
}
}
}
}
}
return 0;
}
2024-05-06T21:50:05 点赞:0