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;

}

pat1041-1050的更多相关文章

  1. [Q&A] MySQL Error 1050(42S01): Table already exist

    [环境说明] 1:MySQL Server 5.5 2:MyEclipse 2014 3:JDK 1.7 造成该问题的可能原因: 1:用 Java 读取 SQL 文件,并执行其中的 sql 语句,但是 ...

  2. ytu 1050:写一个函数，使给定的一个二维数组（３×３）转置，即行列互换（水题）

    1050: 写一个函数,使给定的一个二维数组(3×3)转置,即行列互换 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 154  Solved: 112[ ...

  3. magento -- 解决magento错误：ERROR: Base table or view already exists: 1050 Table ... already exists

    相信有更新magento或者,备份转移magento站点的时候可能会碰到类似这样的错误提示: Base table or view already exists: 1050 Table ... alr ...

  4. --hdu 1050 Moving Tables(贪心)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1050 AC code: #include<stdio.h> #include<str ...

  5. hdu 1050 Moving Tables 解题报告

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1050 这道题目隔了很久才做出来的.一开始把判断走廊有重叠的算法都想错了.以为重叠只要满足,下一次mov ...

  6. NBOJv2 1050 Just Go（线段树/树状数组区间更新单点查询）

    Problem 1050: Just Go Time Limits:  3000 MS   Memory Limits:  65536 KB 64-bit interger IO format:  % ...

  7. PAT 解题报告 1050. String Subtraction (20)

    1050. String Subtraction (20) Given two strings S1 and S2, S = S1 - S2 is defined to be the remainin ...

  8. BZOJ 1050 旅行comf(枚举最小边-并查集)

    题目链接:http://61.187.179.132/JudgeOnline/problem.php?id=1050 题意:给出一个带权图.求一条s到t的路径使得这条路径上最大最小边的比值最小? 思路 ...

  9. hiho #1050 : 树中的最长路 树的直径

    #1050 : 树中的最长路 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 上回说到,小Ho得到了一棵二叉树玩具,这个玩具是由小球和木棍连接起来的,而在拆拼它的过程中, ...

  10. 51nod 1050 循环数组最大子段和

    题目链接:51nod 1050 循环数组最大子段和 #include<stdio.h> #include<algorithm> using namespace std; ; l ...

随机推荐

  1. 前端-Useful Js Plugins

    Validform.min.js:提供对表单的验证.提交等功能,具体可查阅相关文档,@Validform 示例: $("#id").Validform() ; WdatePicke ...

  2. Android Studio,使用外部模拟器作为生成app调试的模拟器

    本文作为一个原理探究的存在. 1.作者观察发现,逍遥和夜神模拟器自身,都是有一个adb.exe和nox_adb.exe存在的,这个东西是干嘛用的呢. 应该是和本身模拟器进行通信的,我测试可以通过他们自 ...

  3. 测试人员如何使用Git部署测试环境

    Git是分布式的版本控制系统. 作为一名Git的小白使用者,一开始接触很懵逼,因为总担心自己一不小心误操作影响代码仓库的代码,网络上关于Git的使用多从开发的角度,很少有人从测试的角度来介绍Git的使 ...

  4. 浅显总结ASCII Unicode UTF-8的区别

    如果觉得此地排版不好,欢迎访问我的博客 浅显总结ASCII Unicode UTF-8的区别 制作表单时,为了追求更好的用户交互体验,常常会有提示性的内容,比如提醒用户字符的限制.由于英文,中文字符的 ...

  5. IDEA的Maven依赖如何引入到External Libraries中

    现象 在Apollo项目中,遇到了一个问题.当在Module的pom.xml中引入依赖: <dependency> <groupId>com.ctrip.framework.a ...

  6. Install Centrifugo and quick start

    Install Centrifugo and quick start Go is a perfect language - it gives developers an opportunity to ...

  7. CentOS安装JDK 8

    准备工作 首先,更新包: yum update 检查服务器上是否已安装旧版本的Java: java -version 如果有旧版本的Java则移除: yum remove java-1.6.0-ope ...

  8. maven项目打包的时候，*Mapper.xml 文件会打不不进去解决办法

    打包的时候,不同版本的 Eclipse 还有IDEA 会有打包打不进去Mapper.xml 文件,这个时候要加如下代码, 在<build> 标签内加入即可 <resources> ...

  9. 对于文件File类型中的目录分隔符

    当我们在使用File f=new File(path); 对于这个path来说是一个很痛苦的事情,在不同的系统上面的分隔符号是不一样的 在windows下面的话是一个"\"而在un ...

  10. Java中Dom解析xml文档

    xml文档 <?xml version="1.0" encoding="UTF-8"?> <bookstore> <book id ...