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\n", A[i]); fl = 1;        break;      }    }    if(!fl) printf("None\n");  }  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("\n");  }  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\n");      postorder(root);        for(int i = 1; i <= n; ++i) {        if(i != 1) printf(" ");          printf("%d", A[output[i]]);      }      printf("\n");      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\n");      postorder(root);      for(int i = 1; i <= n; ++i) {        if(i != 1) printf(" ");        printf("%d", A[output[i]]);      }      printf("\n");      }else printf("NO\n");  }  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\n", 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\n", 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\n", 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\n", 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\n", 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\n", 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\n", re, re); fl = 1;          break;        }      }else {        if(mp.find(re) != mp.end()) {          printf("%d %d\n", coin[i], re); fl = 1;          break;        }      }     }    if(!fl) printf("No Solution\n");  }  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\n", 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\n", 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("\n");  return 0;}