P1897 电梯里的爱情
题目链接：https://www.luogu.com.cn/problem/P1897
#include<iostream> #include<stdio.h> #include<math.h> #include<algorithm> #include<string> using namespace std; int a[100001]; /* 下一个人1s：+ns 每向上运行一层需要 6 秒钟：+h层*6 向下运行一层需要 4 秒钟：+h层*4 每开门一次需要 5 秒：+you层*5 */ int main() { int n,h,you=0,sum=0; cin >> n; for (int i = 0; i < n; i++) { cin >> a[i]; } sort(a, a + n); int cnt = 0; for (int i = 0; i < n; i++) { if (a[i]==a[i+1]) { continue; } else { you++; } } sum = n + a[n - 1] * 10 + you * 5; cout << sum; return 0; }
P1428 小鱼比可爱
题目链接：https://www.luogu.com.cn/problem/P1428
#include<iostream> #include<stdio.h> #include<math.h> #include<algorithm> #include<string> using namespace std; int a[101], b[101]; int main() { int n; cin >> n; for (int i = 0; i < n; i++) { cin >> a[i]; } // 挨个比 for (int i = 0; i < n; i++) { for (int j = i; j >= 0; j--) { if (a[j] < a[i]) b[i]++; } } for (int i = 0; i < n-1; i++) cout << b[i] << " "; cout << b[n-1]; return 0; }
P2676 [USACO07DEC]Bookshelf B
题目链接：https://www.luogu.com.cn/problem/P2676
#include<iostream> #include<stdio.h> #include<math.h> #include<algorithm> #include<string> using namespace std; int h[20000]; int main() { int n,b,cnt=0; cin >> n>>b; for (int i = 0; i < n; i++) { cin >> h[i]; } sort(h,h+n); int stmp = 0; for (int i = n-1; i > 0; i--) { stmp += h[i]; cnt++; if (stmp>b) { break; } } cout << cnt; return 0; }
P4414 [COCI2006-2007#2] ABC
题目链接：https://www.luogu.com.cn/problem/P4414
#include<iostream> #include<stdio.h> #include<math.h> #include<algorithm> #include<string> using namespace std; int a[3]; char c[3]; int main() { cin >> a[0] >> a[1] >> a[2]; cin >> c[0] >> c[1] >> c[2]; sort(a, a + 3); cout << a[c[0] - 'A'] << " " << a[c[1] - 'A'] << " " << a[c[2] - 'A']; return 0; }
P2637 第一次，第二次，成交
题目链接：https://www.luogu.com.cn/problem/P2637
#include<iostream> #include<stdio.h> #include<math.h> #include<algorithm> #include<string> using namespace std; int pr[1001]; int main() { int m, n,price=0,max=0; cin >> n >> m; for (int i = 0; i < n; i++) { cin >> pr[i]; } sort(pr,pr+n); for (int i = n-1; i > 0; i--) { if ((n-i)*pr[i]>max) { max = (n - i)*pr[i]; price = pr[i]; } } cout << price << " " << max; return 0; }
P1226 【模板】快速幂||取余运算
题目链接：https://www.luogu.com.cn/problem/P1226
快速幂函数模板
ll fast_power(ll a, ll b, ll c) { ll ans = 1; a %= c; while (b) { // b 指数是奇数 if (b & 1) { ans = (ans * a) % c; // 指数加1 } a = (a * a) % c; b >>= 1; } return ans; }
快速幂函数模板原理
看这个视频自学，整理算法没劲，懒得再写了：
https://www.bilibili.com/video/BV12r4y1w7tx
【C++/算法】快速幂算法详解
视频中代码：
// 降幂 ll fast_power(ll a,ll b, ll c) { ll ans = 1; while (b) { // b 指数是奇数 if (b%2==1) { ans = ans*a; // 指数加1 a = a * a; b / 2; } else // b是偶数 { a = a * a;; b / 2; } } return ans; } // 对每一步都取余 ll fast_power(ll a, ll b, ll c) { ll ans = 1; while (b) { // b 指数是奇数 if (b % 2 == 1) { ans = (ans * a)%c; // 指数加1 } a = (a * a)%c; b / 2; } return ans; } // 最终 ll fast_power(ll a, ll b, ll c) { ll ans = 1; a %= c; while (b) { // b 指数是奇数 if (b & 1) { ans = (ans * a) % c; // 指数加1 } a = (a * a) % c; b >>= 1; } return ans; }
