pat1031-1040

1031

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<cmath>#include<queue>using namespace std;const int INF = 0x3f3f3f3f;const int N = 1e6+5;const int M = 3e5+5;char s[85];int n1,n2,n3;char mp[100][100];int main() {  while(~scanf("%s", s)) {    memset(mp, 0, sizeof(mp));    int len = strlen(s);    int t1 = len-2;    int t2 = t1/3; int t3 = t1%3;    n1 = n2 = n3 = t2;    if(t3 == 1) n2 ++;    else if(t3 == 2) n1 ++, n3 ++;  //  printf("%d %d %d\n", n1, n2, n3);    int x = 1, y = 1; int cn = 0;    for(int i = 1; i <= n1+1; ++i) {      mp[x][y] = s[cn++];      x ++;    }    x--; y++;      for(int i = 1; i <= n2+1; ++i) {      mp[x][y] = s[cn++];      y ++;    }    y--; x--;    for(int i = 1; i <= n3; ++i) {      mp[x][y] = s[cn++];      x --;    }    for(int i = 1; i <= n1+1; ++i) {      for(int j = 1; j <= n2+2; ++j) {        if(!mp[i][j]) printf(" ");        else printf("%c", mp[i][j]);      }      printf("\n");    }  }  return 0;}

1032 这题如果是 字符相等不一定是后缀，必须指针相等

#include <cstdio>#include <cstring>#include <vector>#include <iostream>#include <queue>using namespace std;const int N = 1e6+5;char Data[N]; int nx[N];vector<char> ans[2];vector<int> indx[2];void dfs(int x, int tag) {  if(x == -1) return;  ans[tag].push_back(Data[x]);  indx[tag].push_back(x);  dfs(nx[x], tag);}int main() {  int s, t, n;  while(~scanf("%d %d %d", &s, &t, &n)) {    ans[0].clear(); ans[1].clear();      indx[0].clear(); indx[1].clear();    for(int i = 1; i <= n; ++i) {      int a; char b; int c;       scanf("%d %c %d", &a, &b, &c);      Data[a] = b; nx[a] = c;    }    dfs(s, 0);    dfs(t, 1);    int Ans = -1;    for(int i = ans[0].size()-1, j = ans[1].size()-1; i >= 0 && j >= 0; --j, --i) {      if(indx[0][i] != indx[1][j]) {        break;      }else {        Ans = indx[0][i];      }    }     if(Ans == -1) printf("-1\n");    else printf("%05d\n", Ans);  }  return 0;}

1033这题有人水的吧！一点不好做。这个贪心我是想了一晚上，具体看代码吧，我无力吐槽,x是上次加油的点，premm和mm是这次我上次加油的点会加多少油的区间

#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 = 505;typedef long long ll;struct Node{    double p; int d;}E[N];int cmp(Node a, Node b) {  return a.d < b.d;}int main() {  int cap, dis, v, n;    while(~scanf("%d %d %d %d", &cap, &dis, &v, &n)) {    for(int i = 1; i <= n; ++i) {      scanf("%lf %d", &E[i].p, &E[i].d);    }    E[0].p = INF; E[0].d = 0;    sort(E+1, E+n+1, cmp);    if(E[1].d != 0) {      printf("The maximum travel distance = 0.00\n");      continue;    }    int x = 0;    int premm = 0;    int mm = 0;  double ans = 0;    while(1) {      int fr = x+1, to;  //    printf("%d %d %d %.2f\n", x, premm, mm, ans);      for(int i = x+1; i <= n; ++i) {        if(E[i].d <= mm) {          to = i;         }else break;      }      double minn = INF*1.0; int minp = 0;      for(int i = fr; i <= to; ++i) {        if(E[i].p < E[x].p) {          minp = i; minn = E[i].p;          break;        }      }  //    printf("%d to %d: %d\n", fr, to, minp);        if(!minp) {        for(int i = fr; i <= to; ++i) {          if(minn > E[i].p) {            minp = i; minn = E[i].p;          }        }    //    printf("chpos: %d\n", minp);        if(mm >= dis) {          ans += (dis - premm)*1.0 /v *E[x].p;          break;        }else if(!minp) break;        else {    //      printf("hh %d %d\n", E[minp].d, E[x].d);          ans += (mm - premm)*1.0 /v * E[x].p;          premm = mm;          mm += E[minp].d -E[x].d;          x = minp;    //      printf("%d\n", mm);        }      }else {        ans += (E[minp].d - premm)*1.0 /v *E[x].p;        premm = E[minp].d;        x = minp;        mm = E[minp].d + cap * v;      }    }    if(mm < dis) printf("The maximum travel distance = %.2f\n", mm*1.0 );    else printf("%.2f\n", ans);  }  return 0;}

1034 并查集一下，无法理解正确率和上一题一样，我觉得我做到现在最难的是1033

#include<iostream>#include<algorithm>#include<cstring>#include<vector>#include<time.h>#include<stdlib.h>#include<map>#include<cstdio>using namespace std;const int N = 2e3+5;#define mp(A,B) make_pair(A,B)map<string, int> mp;char name[N][10];int tot = 0;int pre[N];int fa(int x) {  return pre[x] == x? x: pre[x] = fa(pre[x]);}int cnt[N];vector<int> gang[N];struct Node{  char nam[5];  int per;}E[N];int alltot;int cmp(Node a,Node b) {  for(int i = 0; i < 3; ++i) {    if(a.nam[i] != b.nam[i]) {      return a.nam[i] < b.nam[i];    }  }}int main() {  int n, k;  while(~scanf("%d %d", &n, &k)) {    alltot = 0;    memset(cnt, 0, sizeof(cnt));    tot = 0; mp.clear();    for(int i = 1; i <= 2*n; ++i) pre[i] = i;    for(int i = 1; i <= 2*n; ++i) gang[i].clear();    for(int i = 0; i < n; ++i) {      char a[10]; char b[10]; int c;      scanf("%s %s %d", a, b, &c);          if(mp.find(a) == mp.end()) {        tot ++; mp[a] = tot;        for(int j = 0; j < 3; ++j) name[tot][j] = a[j];      }      if(mp.find(b) == mp.end()) {        tot ++; mp[b] = tot;        for(int j = 0; j < 3; ++j) name[tot][j] = b[j];      }      int a1 = mp[a]; int b1 = mp[b];      int t1 = fa(a1); int t2 = fa(b1);            if(t1 != t2) pre[t1] = t2;      cnt[a1] += c; cnt[b1] += c;      }    for(int i = 1; i <= tot; ++i) {      gang[fa(i)].push_back(i);    }    for(int i = 1; i <= tot; ++i) {      if(gang[i].size() > 2) {        int maxx = -1; int maxp;        int all = 0;        for(int j = 0; j < gang[i].size(); ++j) {          if(maxx < cnt[gang[i][j]] ) {            maxx = cnt[gang[i][j]]; maxp =  j;          }          all += cnt[gang[i][j]];        }        all /= 2;        if(all <= k) continue;        alltot ++;        E[alltot].per = gang[i].size();        for(int j = 0; j < 3; ++j) {          E[alltot].nam[j] = name[gang[i][maxp]][j];        }      }    }    if(alltot == 0) {      printf("0\n"); continue;    }    sort(E+1, E+alltot+1, cmp);    printf("%d\n", alltot);    for(int i = 1; i <= alltot; ++i) {      for(int j = 0; j < 3; ++j) printf("%c", E[i].nam[j]); printf(" ");      printf("%d\n", E[i].per);    }  }  return 0;}

1035 there is /are 注意下

#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 = 505;const int INF = 0x3f3f3f3f;#define MP(x, y) make_pair(x, y)vector<pair<string, string> > vc;int main() {  int n;  while(~scanf("%d", &n)) {    vc.clear();    for(int i = 0; i < n; ++i) {      char a[20]; char b[20];         scanf("%s %s", a, b);      int len = strlen(b);      int fl = 0;      for(int j = 0; j < len; ++j) {        if(b[j] == '1') {          fl = 1; b[j] = '@';        }else if(b[j] == '0') {          fl = 1; b[j] = '%';          }else if(b[j] == 'l') {          fl = 1; b[j] = 'L';        }else if(b[j] == 'O') {          fl = 1; b[j] = 'o';        }      }        if(fl) vc.push_back(MP(a, b));    }    if(vc.size() == 0) {      if(n == 1) printf("There is %d account and no account is modified\n", n);      else       printf("There are %d accounts and no account is modified\n", n);    }else {      printf("%d\n", vc.size());      for(int i = 0; i < vc.size(); ++i) {        printf("%s %s\n", vc[i].first.c_str(), vc[i].second.c_str());      }    }  }  return 0;}

1036

#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)struct Node{  char name[20];  char sex;  char Id[20];  int grade;}E[N];int main() {  int n;  while(~scanf("%d", &n)) {    int minn = INF; int minp = 0;    int maxn = -1; int maxp = 0;    for(int i = 1; i <= n; ++i) {      scanf("%s %c %s %d", E[i].name, &E[i].sex, E[i].Id, &E[i].grade);      if(E[i].sex == 'M') {        if(E[i].grade < minn) {          minn = E[i].grade; minp = i;        }      }else {        if(E[i].grade > maxn) {          maxn = E[i].grade; maxp = i;        }      }    }    if(!maxp) printf("Absent\n");    else printf("%s %s\n", E[maxp].name, E[maxp].Id);    if(!minp) printf("Absent\n");    else printf("%s %s\n", E[minp].name, E[minp].Id);    if(!minp || !maxp) printf("NA\n");    else printf("%d\n", E[maxp].grade - E[minp].grade);  }  return 0;}

1037

#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];int B[N];int main() {  int n, p;  while(~scanf("%d",  &n)) {    for(int i = 1; i <= n; ++i) scanf("%d", &A[i]);    scanf("%d", &p);    for(int i = 1; i <= n; ++i) scanf("%d", &B[i]);    sort(A+1, A+n+1);    sort(B+1, B+n+1);    ll ans = 0;    for(int i = 1; i <= n; ++i) {      ll tmp = 1ll *A[i] *B[i];      if(tmp > 0) ans += tmp;    }    printf("%lld\n", ans);  }  return 0;}

1038 我瞎想的排序方法，不过觉得很有道理

#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)struct Node{  char num[10];}s[N];int cmp(Node A, Node B) {  char a[20]; char b[20];  int cnt;  int l1 = strlen(A.num); int l2 = strlen(B.num);  cnt = 0;  for(int i = 0; i < l1; ++i) a[cnt++] = A.num[i];  for(int i = 0; i < l2; ++i) a[cnt++] = B.num[i];  cnt = 0;  for(int i = 0; i < l2; ++i) b[cnt++] = B.num[i];  for(int i = 0; i < l1; ++i) b[cnt++] = A.num[i];  int l = strlen(a);  for(int i = 0; i < l; ++i) {    if(a[i] != b[i])        return a[i] < b[i];  }  return 1;}int main() {  int n;  while(~scanf("%d", &n)) {    for(int i = 1; i <= n; ++i) {      scanf("%s", s[i].num);    }    sort(s+1, s+n+1, cmp);    int fl = 0;    for(int i = 1; i <= n; ++i) {      for(int j = 0; s[i].num[j]; ++j) {        if(s[i].num[j] != '0') {          fl = 1; printf("%c", s[i].num[j]);        }else if(fl) {          printf("0");        }      }    }    if(!fl) printf("0");    printf("\n");    }  return 0;}

1039 加了一点小优化卡过去的，有大神能进100ms吗，我才170，

#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 n, k;set<int> stu[N];char nam[N][20];map<int, int> ask;vector<int> vc[3000];set<int> ::iterator it;int change(char a[]) {//  for(int i = 0; i < 4; ++i) printf("%c", a[i]); printf("\n");  int tt = 0;  for(int i = 0; i < 3; ++i) {    tt = tt*100+ a[i]-'A';  }  tt = tt*100 + 30+a[3]-'0';  return tt;}int main() {//  printf("%d\n", change("ZZZ9"));  while(~scanf("%d %d", &n, &k)) {  //  ask.clear();  //  for(int i = 1; i <= k; ++i) vc[i].clear();  //  for(int i = 1; i <= n; ++i) stu[i].clear();    for(int i = 1; i <= k; ++i) {      int a,b; scanf("%d %d", &a, &b);      char s[10];         for(int j = 0; j < b; ++j) {        scanf("%s" , s);        vc[a].push_back(change(s));      }          }    for(int i = 1; i <= n; ++i) {      scanf("%s", nam[i]);      ask[change(nam[i])] = i;    }    for(int i = 1; i <= k; ++i) {      for(int j = 0; j < vc[i].size(); ++j) {        int tt = vc[i][j];        if(ask.find(tt) != ask.end()) {          stu[ask[tt]].insert(i);        }      }    }    for(int i = 1; i <= n; ++i) {      printf("%s %lu", nam[i], stu[i].size());    //  sort(stu[i].begin(), stu[i].end());      for(it = stu[i].begin(); it != stu[i].end(); ++it) {        printf(" %d", *it);      //  int tt = stu[i][j];      //  printf(" %d", tt);      }      printf("\n");    }    }  return 0;}

1040 manacher 搞一搞

#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 = 1e4+5;const int INF = 0x3f3f3f3f;#define MP(x, y) make_pair(x, y)char s[N];char Ma[N*2];int Mp[N*2];void Manacher() {  int len = strlen(s);  int l = 0;  Ma[l++] = '$';  Ma[l++] = '#';  for(int i = 0; i < len; ++i) {    Ma[l++] = s[i];    Ma[l++] = '#';  }  Ma[l] = 0;  int mx = 0, id = 0;  for(int i = 0; i < l; ++i) {    Mp[i] = mx>i? min(Mp[2*id-1], mx-i) : 1;    while(Ma[i+Mp[i]] == Ma[i-Mp[i]]) Mp[i] ++;    if(i + Mp[i] > mx) {      mx = i + Mp[i];      id = i;    }  }}   int main() {  gets(s);  int len = strlen(s);  Manacher();  int ans = 0;  for(int i = 0; i < 2*len+2; ++i) {    ans = max(ans, Mp[i]-1);  }  printf("%d\n", ans);  return 0;}