武汉工程大学第三届ACM程序设计新生赛（多校联赛）（线上赛）（热身赛）题解

题目链接：https://ac.nowcoder.com/acm/contest/9162#question
邀请码：acmwitedu2020

A. 最短逃生距离

当输入和输出的数据总数超过1000个时，建议不直接用C++的输入输出以免超时。
各种写法

# include <stdio.h>
# include <math.h>int main() {int n, m;scanf("%d%d", &n, &m);while (m--) {int x, y;scanf("%d%d", &x, &y);printf("%d\n", abs(x - y));}return 0;
}

第二种：如果不懂原理慎用

# include <iostream>
# include <cmath>int main() {std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);int n, m;std::cin >> n >> m;while (m--) {int x, y;std::cin >> x >> y;std::cout << abs(x - y) << "\n";}return 0;
}

第三种：手写输入输出

# include <iostream>
# include <cstdio>
# include <cmath>template <typename T>
void read(T &val) { T x = 0; int bz = 1; char c; for (c = getchar(); (c<'0' || c>'9') && c != '-';  c = getchar()) ; if (c == '-') { bz = -1; c = getchar(); }for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48; val = x * bz;
}template <typename T>
void put(T x){  static char ss[20];  int bas;  if(x < 0) {  putchar('-');  x = -x;  }  if(x == (T)(0)) {  putchar('0');  return;  }  bas = 0;  for(;x;x/=10) ss[bas++] = x % 10 + '0';  for(;bas--;) putchar(ss[bas]);
}  int main() {int n, m;read(n);read(m);while (m--) {int x, y;read(x);read(y);put(abs(x - y));puts("");}return 0;
}

第四种：直接读取数组（如果不懂原理慎用）

# include <iostream>
# include <cstdio>
# include <cmath>const int BUFSIZE = 20 << 20; //40M
char Buf[BUFSIZE + 1], *buf = Buf;template<class T>
void read(T &a) {int sgn = 1;for(a=0; *buf<'0'||*buf>'9'; buf++) if(*buf=='-') sgn = -1;while(*buf>='0'&&*buf<='9') {a=a*10+(*buf-'0');buf++;}a *= sgn;
}void put(int x) {static char s[20];int bas;if(x < 0) {putchar('-');x = -x;}if(x == 0) {putchar('0');return;}bas = 0;for(; x; x/=10)s[bas++] = x%10+'0';for(; bas--;)putchar(s[bas]);
}int main() {fread(Buf, 1, BUFSIZE, stdin);int n, m;read(n);read(m);while (m--) {int x, y;read(x);read(y);put(abs(x - y));printf("\n");}return 0;
}

Java写的时候要IO优化

import java.util.*;
import java.io.*;public class Main {public static void solve() {int T = nextInt();T = nextInt();while (T-- > 0) {long a = nextLong();long b = nextLong();out.println(Math.abs(a - b));}}  public static void main(String[] args) {reader = new BufferedReader(new InputStreamReader(System.in));tokenizer = null;out = new PrintWriter(System.out);solve();out.close();}static BufferedReader reader;static StringTokenizer tokenizer;static PrintWriter out;static int nextInt(){return Integer.parseInt(next());}static long nextLong(){return Long.parseLong(next());}static double nextDouble(){return Double.parseDouble(next());}static String next(){while (tokenizer == null || !tokenizer.hasMoreTokens()) {try {tokenizer = new StringTokenizer(reader.readLine());} catch (IOException e) {throw new RuntimeException(e);}}return tokenizer.nextToken();}
}

Python没有IO优化的方式，因此输入数据多的时候基本不考虑Python。事实上，打ACM基本上都是用的C++，Java和Python写算法题很容易超时。个人观点，不一定正确

B. spj

这题没有啥好说的，举三个例子，还有其他的写法，随便选一种（出题人写special judge时忘记验证换行，不换行也放过了。这题不是我出的，甩锅。。。验题人没有验出问题来应该由验题人背锅？emmmm）

# include <cstdio>int main() {printf("1\n\kcxz");return 0;
}

# include <cstdio>int main() {printf("2\n\小少爷\n宇少");return 0;
}

# include <cstdio>int main() {printf("2\n宇少\n小少爷");return 0;
}

C. 突然想到一个算法

看见水群里有人斗图时发了这张图片，所以就拿这张图片当热身题。
方法一： a b ∗ c = ( a b ) c a^{b*c}={(a^b)}^c ab∗c=(ab)c,每一步都快速幂即可。复杂度 O ( l o g ( n ! ) ) ≈ O ( n l o g ( n ) ) O(log(n!))≈O(nlog(n)) O(log(n!))≈O(nlog(n)) (斯特林公式)

# include <stdio.h>const int mod = 1e9 + 7;int pow(int x, int p) {int ans = 1 % mod;while (p) {if (p & 1) {ans = 1ll * ans * x % mod;}x = 1ll * x * x % mod;p >>= 1;}return ans;
}void solve() {int n;scanf("%d", &n);int ans = n;for (int i = 2; i <= n; i++) {ans = pow(ans, i);}printf("%d\n", ans);
}int main() {int T;scanf("%d", &T);while (T--) {solve();}return 0;
}

方法二：欧拉函数

p p p为质数，所以 n n n和 p p p互质，让阶乘模质数的欧拉函数，再快速幂。复杂度 O ( n ) O(n) O(n)

# include <stdio.h>const int mod = 1e9 + 7;
int phi;int qpow(int x, int p) {int ans = 1 % mod;while (p) {if (p & 1) {ans = 1ll * ans * x % mod;}x = 1ll * x * x % mod;p >>= 1;}return ans;
}int fphi(int n) {int ans = n;for (int i = 2; i <= n / i; i++) {if (n % i == 0) {ans = ans / i * (i - 1);while (n % i == 0) {n /= i;}}}if (n > 1) {ans = ans / n * (n - 1);}return ans;
}void solve() {int n;scanf("%d", &n);int ans = 1;for (int i = 2; i <= n; i++) {ans = 1ll * ans * i % phi;}printf("%d\n", qpow(n, ans));
}int main() {int T;scanf("%d", &T);phi = fphi(mod);while (T--) {solve();}return 0;
}

D. 兀兀的请求

c o s ( π ) = − 1 cos(\pi)=-1 cos(π)=−1， a r c c c o s ( − 1 ) = π arcccos(-1)=\pi arcccos(−1)=π。C语言的库函数里面 a r c c o s ( ) arccos() arccos() 写作 a c o s ( ) acos() acos()。部分编译器 a c o s ( − 1 ) acos(-1) acos(−1)无法通过，需要写成 a c o s ( − 1.0 ) acos(-1.0) acos(−1.0)

# include <stdio.h>
# include <math.h>typedef long long ll;int main() {double ans = acos(-1.0);ans = ans * 100;for (int i = 3; i <= 14; i++) {ans = ans * 10;ll t = ans;printf("%lld", t % 10);}return 0;
}

用参考公式算到第14位极有可能会出现问题