pat1041-1050

没想到半天就做完了10题 = =，这几题太简单了,基本10分钟一题
1041

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y)

int A[N];
map<int, int> mp;
int main() {
  int n;
  while(~scanf("%d", &n)) {
    mp.clear();
    for(int i = 1; i <= n; ++i) scanf("%d", &A[i]), mp[A[i]] ++;

    int fl = 0;
    for(int i = 1; i <= n; ++i) {
      if(mp[A[i]] == 1) {
        printf("%d
", A[i]); fl = 1;
        break;
      }
    }
    if(!fl) printf("None
");
  }
  return 0;
}

1042

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 4e4+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y)

int card[2][N]; int fl;
void put(int x) {
  int t1 = (x-1)/13;
  int t2 = (x-1)%13;
  t2 ++;

  if(t1 == 4) printf("J%d", t2);
  else if(t1 == 3) printf("D%d", t2);
  else if(t1 == 2) printf("C%d", t2);
  else if(t1 == 1) printf("H%d", t2);
  else printf("S%d", t2);
}
int Swap[60];
int main() {
  int K;
  while(~scanf("%d", &K)) {
    fl = 0;
    for(int i = 1; i <= 54; ++i) card[fl][i] = i;

    for(int i = 1; i <= 54; ++i) scanf("%d", &Swap[i]);

    for(int i = 0; i < K; ++i) {
      for(int j = 1; j <= 54; ++j) {
        card[fl^1][Swap[j]] = card[fl][j];
      }
      fl ^= 1;
    }

    for(int i = 1; i <= 54; ++i) {
      if(i != 1) printf(" ");
      put(card[fl][i]);
    }
    printf("
");
  }
  return 0;
}

1043 这题还是花了一点时间。这是一个反证思想，虽然你按照这个序列插入的树，其他序列可能插入也是这样，但是你反过来验证一下不就行了= =

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e3+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y)

int n;
int A[N];
int lef[N], righ[N];
int root;


void find(int x, int id) {
  if(A[x] <= A[id]) {
    if(righ[x] == 0) righ[x] = id;
    else find(righ[x], id);
  }else {
    if(lef[x] == 0) lef[x] = id;
    else find(lef[x], id);
  }
} 

void _find(int x, int id) {
  if(A[x] > A[id]) {
    if(righ[x] == 0) righ[x] = id;
    else _find(righ[x], id);
  }else {
    if(lef[x] == 0) lef[x] = id;
    else _find(lef[x], id);
  }
} 
int output[N]; int cnt = 0;
void preorder(int x) {
  ++cnt; output[cnt] = x;
  if(lef[x])  preorder(lef[x]);
  if(righ[x]) preorder(righ[x]);
}
void postorder(int x) {
  if(lef[x]) postorder(lef[x]);
  if(righ[x]) postorder(righ[x]);
  ++cnt; output[cnt] = x;
}

int main() {
  while(~scanf("%d", &n)) {
    memset(lef, 0, sizeof(lef));
    memset(righ, 0, sizeof(righ));

    for(int i = 1; i <= n; ++i) {
      scanf("%d", &A[i]);  
    }

    root = 1;
    for(int i = 2; i <= n; ++i) {
      find(root, i);
    }
    cnt = 0;
    preorder(root);
    int flag = 1;
    for(int i = 1; i <= n; ++i) {
      if(output[i] != i) {
        flag = 0; break;
      }    
    }

    if(flag) {
      cnt = 0;
      printf("YES
");
      postorder(root);  
      for(int i = 1; i <= n; ++i) {
        if(i != 1) printf(" ");
          printf("%d", A[output[i]]);
      }
      printf("
");
      continue;
    }

    memset(lef, 0, sizeof(lef));
    memset(righ, 0, sizeof(righ));

    root = 1;
    for(int i = 2; i <= n; ++i) {
      _find(root, i);
    }
    cnt = 0;
    preorder(root);
    flag = 1;
    for(int i = 1; i <= n; ++i) {
      if(output[i] != i) {
        flag = 0; break;
      }    
    }

    if(flag) {
      cnt = 0;
      printf("YES
");
      postorder(root);
      for(int i = 1; i <= n; ++i) {
        if(i != 1) printf(" ");
        printf("%d", A[output[i]]);
      }
      printf("
");  
    }else printf("NO
");




  }
  return 0;
}

1044 二分一下就行，比较简单

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<algorithm>
#include<ctime>
#include<cstdlib>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5+5;
typedef long long ll;

int value[N];
int suf[N];

int main() {
  int n, m;
  while(~scanf("%d %d", &n, &m)) {
    int fl = 0;
    for(int i = 1; i <= n; ++i) {
      scanf("%d", &value[i]);
  //    if(value[i] <= 0) while(1);
    }

    suf[0] = 0;

    int pre = 0;
    int minn = INF;
    for(int i = 1; i <= n; ++i) {
      pre += value[i];
      suf[i] = pre;
  //    printf("%d ", suf[i]);
    //  if(i == 1) continue;

      int id = lower_bound(suf, suf+i, pre-m) - suf;
      if(suf[id] == pre-m) {
        printf("%d-%d
", id+1, i);
        fl = 1;
      }else {
        if(id == 0) continue;
        int tt = suf[i] - suf[id-1];
        if(tt > m) minn = min(minn, tt);
      }
    }

    if(!fl) {
      pre = 0;
      for(int i = 1; i <= n; ++i) {
        pre += value[i];
        suf[i] = pre;
      //  if(i == 1) continue;
        int id = lower_bound(suf, suf+i, pre - minn) - suf;
        if(suf[id] == pre - minn) {
          printf("%d-%d
", id+1, i);
        //  fl = 1;
        }
      }
    }
  }
  return 0;
}

1045 这题dp，每次讨论一个点，找到此时，小于等于该物品喜欢程度为结尾的物品的最大值，当然我用树状数组加速了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<algorithm>
#include<ctime>
#include<cstdlib>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5+5;
typedef long long ll;

int fav[205];
int fav_[205];
int stripe[10005];

int tree[205];
int n, m, len;
void add(int pos, int num) {
  for(int i = pos; i <= m; i += i&-i) {
    tree[i] = max(tree[i], num);  
  }
}
int sum(int pos) {

  int ans = 0;
  for(int i = pos; i; i -= i&-i) {
    ans = max(ans, tree[i]);
  }
  return ans;
}



int main() {
  while(~scanf("%d", &n)) {
    memset(fav_, 0, sizeof(fav_));
    memset(tree, 0, sizeof(tree));

    scanf("%d", &m);
    for(int i = 1; i <= m; ++i) {
      scanf("%d", &fav[i]);
      fav_[fav[i]] = i;
    }
    scanf("%d", &len);
    for(int i = 1; i <= len; ++i) {
      scanf("%d", &stripe[i]);
    }

    for(int i = 1; i <= len; ++i) {
      if(fav_[stripe[i]] == 0) continue;

      int id = fav_[stripe[i]];
      int tt = sum(id)+1;


      add(id, tt);
    }

    printf("%d
", sum(m));
  }
  return 0;
}

1046

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<algorithm>
#include<ctime>
#include<cstdlib>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5+5;
typedef long long ll;

int d[N];
ll suf[N];
int main() {
  int n, m;
  while(~scanf("%d", &n)) {

    suf[0] = 0;
    for(int i = 1; i <= n; ++i) {
      scanf("%d", &d[i]);
      suf[i] = suf[i-1] + d[i];
    }
    scanf("%d", &m);
    for(int i = 1; i <= m; ++i) {
      int a, b; scanf("%d %d", &a, &b);
      if(a > b) swap(a, b);
      ll tmp = suf[b-1] - suf[a-1];
      ll ans = min(tmp, suf[n] - tmp);
         printf("%lld
", ans);  
    }


  }
  return 0;
}

1047 还是先将字符串处理成了数字，这样更快

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<algorithm>
#include<ctime>
#include<cstdlib>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5+5;
typedef long long ll;

int change(char *a) {
  int ans = 0;
  for(int i = 0; i < 3; ++i) {
    ans = ans*100 + a[i]-'A';
  }
  ans = ans*100 + a[3]-'0'+30;
  return ans;
}
void put(int x) {
  printf("%c%c%c%c
", x/1000000+'A', (x%1000000)/10000+'A', (x%10000)/100+'A', (x%100-30)+'0' ); 
}
vector<int> course[2505];

int main() {
  int n, k;
  while(~scanf("%d %d", &n, &k)) {
    for(int i = 1; i <= k; ++i) course[i].clear();

    for(int i = 0; i < n; ++i) {
      char s[10]; int a;
      scanf("%s %d", s, &a);
      int id = change(s);
      for(int j = 0; j < a; ++j) {
        int b; scanf("%d", &b);
        course[b].push_back(id);
      } 
    }

    for(int i = 1; i <= k; ++i) {
      printf("%d %d
", i, course[i].size());
      sort(course[i].begin(), course[i].end());
      for(int j = 0; j < course[i].size(); ++j) {
        put(course[i][j]);
      }
    }
  }
  return 0;
}

1048

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<algorithm>
#include<ctime>
#include<cstdlib>
#include<map>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5+5;
typedef long long ll;

int coin[N];
map<int, int> mp;
int main() {
  int n, m;
  while(~scanf("%d %d", &n, &m)) {
    mp.clear();
    for(int i = 1; i <= n; ++i) {
      scanf("%d", &coin[i]);
      mp[coin[i]] ++;
    }

    sort(coin+1, coin+n+1);
    int fl = 0;
    for(int i = 1; i <= n; ++i) {
      int re = m - coin[i];

      if(re == coin[i]) {
        if(mp[re] >= 2) {
          printf("%d %d
", re, re); fl = 1;
          break;
        }
      }else {
        if(mp.find(re) != mp.end()) {
          printf("%d %d
", coin[i], re); fl = 1;
          break;
        }
      } 
    }
    if(!fl) printf("No Solution
");
  }
  return 0;
}

1049 这题还是要想下，先预处理 0-999999 那么一个大于1e6的数就可以拆分成前五位，后五位来做就行啦

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<algorithm>
#include<ctime>
#include<cstdlib>
#include<map>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e6;
typedef long long ll;

ll num[N];
int cal(int x) {
  int ans = 0;
  while(x) {
    int tt = x%10;
    if(tt == 1) ans ++;
    x /= 10;  
  }
  return ans;
}
int main() {
  num[0] = 0;
  for(int i = 1; i < N; ++i) {
    num[i] = num[i-1] + cal(i);
  }

  int x;
  while(~scanf("%d", &x)) {
    if(x < 1e6) printf("%lld
", num[x]);
    else {
      int t1 = x % 1000000; int t2 = x / 1000000;
      ll ans = (num[t2]-cal(t2)) * 1e6 + 1ll*cal(t2)*(t1+1) + num[999999] * t2 + num[t1];
      printf("%lld
", ans);
    }
  }
  return 0;
}

1050

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<algorithm>
#include<ctime>
#include<cstdlib>
#include<map>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e4+5;
typedef long long ll;

char a[N], b[N];
int has[200];
int main() {
  memset(has, 0, sizeof(has));
  gets(a); gets(b);

  for(int i = 0; b[i]; ++i) {
    has[b[i]] ++;
  }

  for(int i = 0; a[i]; ++i) {
    if(!has[a[i]]) printf("%c", a[i]);
  }
  printf("
");
  return 0;
}