【HDOJ】2050 - 2059
引言：两年没写博客了，两年来经历了很多的事情，太多太多，无法说清(；′⌒`)。无论如何生活还要继续，当前计划从杭电OJ开始记录我的痕迹(ง •_•)ง。
(如有错误，欢迎指正o(￣▽￣)ブ)
目录
- 2050 (折线分割平面)
- 2051 (Bitset)
- 2052 (Picture)
- 2053 (Switch Game)
- 2054 (A == B ?)
- 2055 (An easy problem)
- 2056 (Rectangles)
- 2057 (A + B Again)
- 2058 (The sum problem)
- 2059 (龟兔赛跑)
2050 (折线分割平面)
#include
using namespace std;
int main() {
int c, n;
cin >> c;
long long b[20001];
b[1] = 2;
for (int i = 2; i <= 20000; i++)
b[i] = b[i - 1] + i;
while (c--) {
cin >> n;
cout << b[2*n - 1] << endl;
}
}
详细分析在这里:【HDOJ】2050 (折线分割平面)
2051 (Bitset)
#include
#include
using namespace std;
//用栈代替了递归的方式
int main() {
int num;
stack outPut;
while (cin >> num) {
while (num > 1) {
outPut.push(num % 2);
num /= 2;
}
if (num == 1) outPut.push(num);
while (!outPut.empty()) {
cout << outPut.top();
outPut.pop();
}
cout << endl;
}
}
2052 (Picture)
#include
using namespace std;
int main() {
int n, m;
while (scanf("%d%d", &n, &m) != EOF) {
cout << "+";
for (int i = 1; i <= n; i++) cout << "-";
cout << "+" << endl;
for (int i = 1; i <= m; i++) {
cout << "|";
for (int j = 1; j <= n; j++) cout << " ";
cout << "|" << endl;
}
cout << "+";
for (int i = 1; i <= n; i++) cout << "-";
cout << "+" << endl;
}
}
2053 (Switch Game)
#include
using namespace std;
//这道题的难点应该是看不太懂英文吧(；′⌒`)
int main() {
int n;
while (cin >> n) {
int res = 0;
for (int i = 1; i <= n; i++) {
if (n % i == 0) res = res == 0 ? 1 : 0;
}
cout << res << endl;
}
}
2054 (A == B ?)
#include
#include
using namespace std;
string dealNum(string num) {
int i = 0, start = 0, len = num.size();
if (len == 0) return "";
//处理符号
string op = "";
if (num[0] == '+' || num[0] == '-') {
i++;
start++;
len--;
if (num[0] == '-') op = "-";
}
//处理整数前面的0
while (i < num.size() && num[i] == '0') {
i++;
start++;
len--;
}
//处理小数点后面的0
bool isPoint = false;
while (i < num.size()) {
if (num[i] == '.') {
isPoint = true;
break;
}
i++;
}
if (isPoint) {
i = num.size() - 1;
while (num[i] == '0') {
i--;
len--;
}
if (num[i] == '.') {
i--;
len--;
}
}
// 处理0
if (len == 0) return "0";
return op + num.substr(start, len);
}
int main() {
string a, b;
while (cin >> a >> b) {
a = dealNum(a);
b = dealNum(b);
cout << (a == b ? "YES" : "NO") << endl;
}
}
2055 (An easy problem)
#include
using namespace std;
int main() {
int len;
cin >> len;
char x;
int y, res;
while (len--) {
cin >> x >> y;
if (x >= 'A' && x <= 'Z') res = x - 'A' + 1 + y;
else if (x >= 'a' && x <= 'z') res = y - (x - 'a' + 1);
cout << res << endl;
}
}
2056 (Rectangles)
#include
#include
using namespace std;
// --------------3 --------------1
// - - - - - -
// - - - - - -
// ------1 - ------3 -
// - 2 --------- - 0 ---------
// - - - - - -
// 0-------------- 2--------------
// 总是把(x1,y1)化成0的位置
// 总是把(x2,y2)化成1的位置
// 总是把(x3,y3)化成2的位置
// 总是把(x4,y4)化成3的位置
// 在矩阵相交的情况下,相交面积始终是第3大的坐标-第2大的坐标
int main() {
double x1, y1, x2, y2, x3, y3, x4, y4;
while (scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3, &x4, &y4) != EOF) {
//处理矩形A的边界
double a_left = min(x1, x2);
double a_right = max(x1, x2);
double a_bottom = min(y1, y2);
double a_top = max(y1, y2);
//处理矩形B的边界
double b_left = min(x3, x4);
double b_right = max(x3, x4);
double b_bottom = min(y3, y4);
double b_top = max(y3, y4);
//计算相交区域的边界
double overlap_left = max(a_left, b_left);
double overlap_right = min(a_right, b_right);
double overlap_bottom = max(a_bottom, b_bottom);
double overlap_top = min(a_top, b_top);
double res = 0;
if (overlap_left < overlap_right && overlap_bottom < overlap_top) {
res = (overlap_right - overlap_left) * (overlap_top - overlap_bottom);
}
printf("%.2lf\n", res);
}
}
2057 (A + B Again)
#include
using namespace std;
//这道题限制了a,b的长度,所以可以用长整形
//如果不限制a,b长度呢,我的方法是:解析字符,模拟计组那一套计算方式
int main() {
long long a, b, res;
//llx表示长整形的16进制形式,llX大写X表示十六进制大写
while (scanf("%llx%llx", &a, &b) != EOF) {
res = a + b;
if (res < 0) {
printf("-%llX\n", -res);
}
else {
printf("%llX\n", res);
}
}
}
2058 (The sum problem)
#include
using namespace std;
//本想着滑动窗口秒了,结果超时了
int main() {
long long n, m;
while (cin >> n >> m) {
if (n == 0 && m == 0) break;
long long left = 1, right = 1, target = 2 * m;
while (left <= n && right <= n && left <= right) {
if (left == m || right == m) {
printf("[%lld,%lld]\n", m, m);
break;
}
long long val = (left + right) * (right - left + 1);
if (val == target) {
printf("[%lld,%lld]\n", left, right);
right++;
}
else if (val < target) right++;
else left++;
}
printf("\n");
}
}
#include
using namespace std;
//优化算法
// (left + right) * (right - left + 1) = 2 * m ---- (首项+末项)*项数/2
// (2left + len - 1) * len = 2 * m ---- 求left, right
// => left = m / len + (1 - len) / 2
// => right = left + len - 1
// ---------------------------------
// (2left + len - 1) * len = 2 * m ---- 求len, left = 1时, len最大
// => (len + 1) * len = 2 * m
// => len * len < 2 * m
// => len < sqrl(2 * m)
int main() {
long long n, m;
while (cin >> n >> m) {
if (n == 0 && m == 0) break;
long long left, right, target = 2 * m;
for (long long len = sqrt(target); len >= 1; len--) {
left = m / len + (1 - len) / 2;
right = left + len - 1;
if ((left + right) * len == target)
printf("[%lld,%lld]\n", left, right);
}
printf("\n");
}
}
2059 (龟兔赛跑)
#include
#include
using namespace std;
//不能AC,因为到达一个充电站后,不同的剩余电量会影响最优路径的选择
//测试用例:
//40
//6 20 5
//5 8 2
//20 24 28 32 36 39
int main() {
int l, n, c, t, vr, vt1, vt2, arr[102];
double dp[102];
while (cin >> l) {
cin >> n >> c >> t >> vr >> vt1 >> vt2;
arr[0] = 0;
for (int i = 1; i <= n; i++) cin >> arr[i];
arr[n + 1] = l;
//乌龟到每一个点耗时
dp[0] = 0;
int rest_c = c;//剩余电力可以跑的距离
for (int i = 1; i <= n + 1; i++) {
int s = arr[i] - arr[i - 1];
//剩余电力能跑到下一个充电站
if (rest_c >= s) {
rest_c -= s;
dp[i] = dp[i - 1] + (double)s / vt1;
}
//剩余电力不能跑到下一个充电站
//充电花的时间,和不充电花的时间,取最小值
else {
double time_no_charge, time_charge;
int new_rest;
//不充电花的时间
time_no_charge = (double)rest_c / vt1 + (double)(s - rest_c) / vt2;
//充满电后能够跑到下一个充电站
if (c >= s) {
new_rest = c - s;
time_charge = (double)s / vt1 + t;
}
//充满电后也不能够跑到下一个充电站
else {
new_rest = 0;
time_charge = (double)c / vt1 + (double)(s - c) / vt2 + t;
}
//不充电花的时间更多
if (time_no_charge > time_charge) {
rest_c = new_rest;
dp[i] = dp[i - 1] + time_charge;
}
//充电花的时间更多
else {
rest_c = 0;
dp[i] = dp[i - 1] + time_no_charge;
}
}
}
if (dp[n + 1] > ((double)l / vr)) cout << "Good job,rabbit!";
else cout << "What a pity rabbit!";
}
}
#include
#include
using namespace std;
// 由于之前最优路径的选择依赖于前一段路径
// 所以可以通过遍历前面所有路径,选出最短时间
int main() {
int l, n, c, t, vr, vt1, vt2, arr[102];
double dp[102];
while (cin >> l) {
cin >> n >> c >> t >> vr >> vt1 >> vt2;
arr[0] = 0;
for (int i = 1; i <= n; i++) cin >> arr[i];
arr[n + 1] = l;
//乌龟到每一个点耗时
dp[0] = 0;
for (int i = 1; i <= n + 1; i++) {
dp[i] = -1;
double time;
//遍历前面每一个充电站,比较在那个充电站充电和不充电时,取花费的最小时间
for (int j = 0; j < i; j++) {
int s = arr[i] - arr[j];
if (c >= s) time = dp[j] + (double)s / vt1;
else time = dp[j] + (double)c / vt1 + (double)(s - c) / vt2;
if (j > 0) time += t;
if (dp[i] == -1) dp[i] = time;
else dp[i] = min(dp[i], time);
}
}
if (dp[n + 1] > ((double)l / vr)) cout << "Good job,rabbit!";
else cout << "What a pity rabbit!";
}
}