2015 Fall HIT Weekly Training 2

在错过了第一次之后，终于准时参加了第二次。

18:30-20:30的新手练习题，我居然真的就做了两个小时，差点没AC

A.YY, and YY again

　　读取string时会停在空格，所以用getline就行。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int main(void)
{
  string st;
  int x;
  while(getline(cin, st)){
    x = 0;
    for(int i = 0; i < st.length(); ++i)
      if(st[i] >= 'A' && st[i] <= 'Z')
    x += st[i] - 'A' + 1;
    if(x > 100) cout<<"INVALID"<<endl;
    else cout<<x<<endl;

  }
  return 0;
}

 

B.Box of Bricks

　　这年头，思路比这再简单的数学题真是不多见了。

     计算∑(a[i]<p)(p-a[i])就行了。p是最后的高度。

#include<cstdio>

int a[100];

int main(void)
{
  int N, k = 1;
  int p;
  int ans;
  scanf("%d", &N);
  while(N > 0){
    if(k > 1) printf("\n");
    p = 0;
    ans = 0;
    for(int i = 0; i < N; ++i){
      scanf("%d", &a[i]);
      p += a[i];
    }
    p /= N;
    for(int i = 0; i < N; ++i)
      if(a[i] < p)
    ans += p - a[i];
    printf("Set #%d\n", k++);
    printf("The minimum number of moves is %d.\n", ans);
    scanf("%d", &N);
  }
  return 0;
}

 

C.Not A Water Problem

　　首先是那个诡异Hint：That isn't a real water problem,but I promise that there's only one trick.

　　一看就知道这道题暗藏杀机，开始时候使用while(scanf())，妥妥的TLE。

　　奥，原来就trick就是EOF啊，果断改，试了!EOF、getchar() != EOF 、(ch = getchar()) != EOF。果断都不管用，花式WA。。。

　　然后恍然大悟，signed int，果然是个暗示啊，abs(-2^31)不能用int，然后换LL，还是各种WA。。。

　　在最后的两分钟，开始各种尝试，终于当我换回了while(cin)的时候，它红了。

　　。。。。。。

#include<cstdio>
#include<iostream>
using namespace std;

#define abs(a) ((a)>0)?(a):(-(a))

int main(void)
{
  long long x;
  while(cin>>x){
    cout<<(abs(x))<<endl;
  }
  return 0;
}

 

D.整数的排序

　　果断的sort

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstdlib>

int a[20];

int main(void)
{
  int N;
  scanf("%d", &N);
  for(int i = 0; i < N; ++i){
    for(int j = 0; j < 10; ++j) scanf("%d", &a[j]);
    std::sort(a, a+10);
    for(int j = 0; j < 9; ++j) printf("%d ", a[j]);
    printf("%d\n", a[9]);
  }
  return 0;
}

 

E.Fibonacci Numbers

　　看见这题瞬间就想起了那年哪月，翔神教我的矩阵快速幂求斐波那契。。。REBMYSMLY

　　偷偷的百度了lrj的bign，ctrl+c， ctrl+p。

　　想着有lrj庇佑，这题肯定一次AC，然而。。。

　　经典坑爹，教科书式的坑爹，测试数据有多组，但他就是不告诉你。

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

#define MAXN 3010

struct bign{
  int len, s[MAXN];

  bign() {
    memset(s, 0, sizeof(s));
    len = 1;
  }

  bign(int num) {
    *this = num;
  }

  bign(const char* num) {
    *this = num;
  }

  bign operator = (int num) {
    char s[MAXN];
    sprintf(s, "%d", num);
    *this = s;
    return *this;
  }

  bign operator = (const char* num) {
    len = strlen(num);
    for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
    return *this;
  }

  string str() const {
    string res = "";
    for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;
    if(res == "") res = "0";
    return res;
  }

  bign operator + (const bign& b) const{
    bign c;
    c.len = 0;
    for(int i = 0, g = 0; g || i < max(len, b.len); i++) {
      int x = g;
      if(i < len) x += s[i];
      if(i < b.len) x += b.s[i];
      c.s[c.len++] = x % 10;
      g = x / 10;
    }
    return c;
  }

  void clean() {
    while(len > 1 && !s[len-1]) len--;
  }

  bign operator * (const bign& b) {
    bign c; c.len = len + b.len;
    for(int i = 0; i < len; i++)
      for(int j = 0; j < b.len; j++)
        c.s[i+j] += s[i] * b.s[j];
    for(int i = 0; i < c.len-1; i++){
      c.s[i+1] += c.s[i] / 10;
      c.s[i] %= 10;
    }
    c.clean();
    return c;
  }

  bign operator - (const bign& b) {
    bign c; c.len = 0;
    for(int i = 0, g = 0; i < len; i++) {
      int x = s[i] - g;
      if(i < b.len) x -= b.s[i];
      if(x >= 0) g = 0;
      else {
        g = 1;
        x += 10;
      }
      c.s[c.len++] = x;
    }
    c.clean();
    return c;
  }

  bool operator < (const bign& b) const{
    if(len != b.len) return len < b.len;
    for(int i = len-1; i >= 0; i--)
      if(s[i] != b.s[i]) return s[i] < b.s[i];
    return false;
  }

  bool operator > (const bign& b) const{
    return b < *this;
  }

  bool operator <= (const bign& b) {
    return !(b > *this);
  }

  bool operator == (const bign& b) {
    return !(b < *this) && !(*this < b);
  }

  bign operator += (const bign& b) {
    *this = *this + b;
    return *this;
  }
};

istream& operator >> (istream &in, bign& x) {
  string s;
  in >> s;
  x = s.c_str();
  return in;
}

ostream& operator << (ostream &out, const bign& x) {
  out << x.str();
  return out;
}


int main(void)
{
  int N;
  bign a1, a2;
  while(cin>>N){
    a1 = 1;
    a2 = 1;
  for(int i = 3; i <= N; i += 2){
    a1 = a1 + a2;
    a2 = a1 + a2;
  } 
  if(N & 1) cout<<a1<<endl;
  else cout<<a2<<endl;
  }
  return 0;
}