pat1011-1020

一开始几道题写到吐血，真的自己现在好弱

1011 水题不说了

#include<bits/stdc++.h>
using namespace std;
const int N = 105;
typedef unsigned long long ll;
const int MOD = 1e9+7;

int c[5];
int main() {
double ans = 1;
for(int i = 1; i <= 3; ++i) {
double mx = -1;
for(int j = 1; j <= 3; ++j) {
double a; scanf("%lf", &a);
if(mx < a) {
mx = a; c[i] = j;
}
}
ans *= mx;
}
ans = (ans*0.65-1)*2;
for(int i = 1; i <= 3; ++i) {
printf("%s ", c[i]==1? "W": (c[i]==2?"T":"L") );
}
printf("%.2f\n", ans);
}

1012 比较麻烦，一开始处理一下

#include<bits/stdc++.h>
using namespace std;
const int N = 105;
typedef unsigned long long ll;
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;

int n, m;
char s[2005][8];
int C[5][2005];
int cc[5][105];
map<string, int> cmp;
map<int,int> mp;
map<int,int> ::iterator it;

int main() {
while(~scanf("%d %d",&n,&m)) {
memset(cc, 0, sizeof(cc));
cmp.clear();

for(int i = 1; i <= n; ++i) {
scanf("%s %d %d %d", s[i], &C[0][i], &C[1][i], &C[2][i]);
cmp[s[i]] = i;

C[3][i] = (C[0][i]+C[1][i]+C[2][i]) / 3;
//iprintf(" %d", C[3][i]);
}
int cnt = 1;

for(int j = 0; j < 4; ++j) {
mp.clear();
for(int i = 1; i <= n; ++i) {
mp[C[j][i]] ++;
}
cnt = 1;
for(it = mp.end(), --it; ; --it) {
cc[j][it->first] = cnt;
cnt += it->second;
if(it == mp.begin()) break;
}
for(int i = 1; i <= n; ++i) {
C[j][i] = cc[j][C[j][i]];
}
}

//  for(int i = 1; i <= n; ++i) {
//      for(int j = 0; j < 4; ++j) printf("%d ", C[j][i]); printf("\n");
//    }

for(int i = 1; i <= m; ++i) {
char tmp[15]; scanf("%s", tmp);
if( cmp.find(tmp) == cmp.end() ) printf("N/A\n");
else {
int tt = cmp[tmp];
int po, mi = INF;
//  for(int j = 0; j < 4; ++j) printf("%d ",C[j][tt]); printf("\n");

if(mi > C[3][tt]) {
mi = C[3][tt]; po = 3;
}
for(int j = 0; j < 3; ++j) {
if(mi > C[j][tt]) {
mi = C[j][tt]; po = j;
}
}

printf("%d ", mi);
if(po == 0) printf("C\n");
else if(po == 1) printf("M\n");
else if(po == 2) printf("E\n");
else printf("A\n");
}
}
}
return 0;
}

1013 dfs一下

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int N = 1e3+5;

int n, m, k;
struct Node{
int nx, to;
}E[N*N*2];
int head
, tot;

void add(int fr,int to) {
E[tot].to = to; E[tot].nx = head[fr]; head[fr] = tot++;
}
int ans
;
int cnt; int nop;
int vis
;
void dfs(int x) {
for(int i = head[x]; ~i; i = E[i].nx) {
int to = E[i].to;
if(!vis[to] && to != nop) {
vis[to] = 1;
dfs(to);
}
}
}
void solve(int x) {
cnt = 0;
memset(vis, 0, sizeof(vis));
nop = x;
for(int i = 1; i <= n; ++i) {
if(i != nop && !vis[i]) {
dfs(i); cnt ++;
}
}
ans[x] = cnt-1;
}

int main() {
while(~scanf("%d %d %d", &n, &m, &k)) {
memset(head, -1, sizeof(head)); tot = 0;
memset(ans, -1, sizeof(ans));

for(int i = 0; i < m; ++i) {
int a, b; scanf("%d %d", &a, &b);
add(a, b); add(b, a);
}

for(int i = 1; i <= k; ++i) {
int a; scanf("%d",&a);
if(ans[a] == -1) solve(a);
printf("%d\n", ans[a]);
}

}
return 0;
}

1014 这道题我的理解完全不一样，对于5点下班这件事，我真的无语，这种坑就需要猜了

#include<bits/stdc++.h>
using namespace std;
const int N = 1e4+5;
const int INF = 0x3f3f3f3f;
typedef long long ll;

int usetime[1005];
int ans[1005];
queue<pair<int,int> > Q[25];

void solve(int id, int x) {
int t1 = (x-usetime[id])/60; int t2 = (x-usetime[id])%60;
if(t1+8 >= 17 ) {
printf("Sorry\n");
return;
}

t1 = x/60; t2 = x%60;
printf("%02d:%02d\n", t1+8, t2);
}

int main() {
int n, m, k, q;
while(~scanf("%d %d %d %d", &n, &m, &k, &q)) {
memset(ans, 0, sizeof(ans));
for(int i = 1; i <= n; ++i) {
if(!Q[i].empty()) Q[i].pop();
}

for(int i = 1; i <= k; ++i) {
scanf("%d", &usetime[i]);
}
for(int i = 1; i <= min(k, n*m); ++i) {
int tt = (i-1)%n + 1;
Q[tt].push(make_pair(usetime[i],i));
}

int ntime = 0;
for(int i = min(k,n*m)+1; i <= k; ++i) {
int mi = INF; int po;
for(int j = 1; j <= n; ++j) {
int ti = Q[j].front().first;
if(mi > ti) {
mi = ti; po = j;
}
}

int t2 = Q[po].front().second; Q[po].pop();
ans[t2] = ntime+mi;
ntime += mi;
Q[po].push(make_pair(usetime[i],i));

for(int j = 1; j <= n; ++j) {
if(j != po) {
Q[j].front().first -= mi;
}
}
}

for(int j = 1; j <= n; ++j) {
int wtime = ntime;
while(!Q[j].empty()) {
int t1 = Q[j].front().first; int t2 = Q[j].front().second; Q[j].pop();
ans[t2] = wtime+t1;
wtime += t1;
}
}

for(int i = 1; i <= q; ++i) {
int a; scanf("%d", &a);
if(ans[a] == 0) while(1);
solve(a, ans[a]);
}
}
return 0;
}

/*

2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7

*/

1015深坑， 1000翻转0001不合法的

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+5;

int prime(int x) {
//  printf("%d\n", x);
for(int i = 2; i < x; ++i) {
if(x % i == 0) {
return 0;
}
}
return 1;
}
int main() {
int n, d;
while(~scanf("%d", &n)) {
if(n < 0) break;
scanf("%d",&d);

int tt = n;
vector<int> vc;
while(tt) {
vc.push_back(tt%d);
tt /= d;
}

int _n = 0;
if(vc[0] == 0) {
printf("No\n"); continue;
}
for(int i = 0; i < vc.size(); ++i) {
//    printf("%d\n",vc[i]);
_n = _n*d + vc[i];
}

if(prime(n) && prime(_n)) printf("Yes\n");
else printf("No\n");
}
return 0;
}

1016 很烦一题，我的世界一开始计算错了

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

int toll[30];
char s[1005][50];
struct Node{
int mon, day, hour, mina; int on;
Node(int a=0, int b=0, int c=0, int d=0, int e=0):mon(a), day(b), hour(c), mina(d), on(e){}
};

map<string, int> mp; int tot;
map<string, int> ::iterator it;
vector<Node> tim[1005];
vector< pair< pair<Node,Node>, pair<int, int> > > ans[1005][15];

int cmp(Node a, Node b) {
if(a.mon != b.mon) return a.mon < b.mon;
else if(a.day != b.day) return a.day < b.day;
else if(a.hour != b.hour) return a.hour < b.hour;
else return a.mina < b.mina;
}

int main() {
int allhour = 0;
for(int i = 0; i <= 23; ++i) {
scanf("%d", &toll[i]);
allhour += toll[i];
}
tot = 0;

int n;
scanf("%d", &n);
for(int i = 1; i <= n; ++i) {
char tmp[10];
scanf("%s", s[i]);
int a,b,c,d, e=1;
scanf("%d:%d:%d:%d", &a, &b, &c, &d);

scanf("%s", tmp);
if(tmp[1] == 'f') e *= -1;

if(mp.find(s[i]) == mp.end()) mp[s[i]] = ++tot;
tim[mp[s[i]]].push_back(Node(a,b,c,d,e));
}

for(int i = 1; i <= tot; ++i) {
sort(tim[i].begin(), tim[i].end(), cmp);
for(int j = 0; j < tim[i].size(); j += 2) {
if(tim[i][j].on > 0 && j+1 < tim[i].size() && tim[i][j+1].on < 0) {
int a1, a2;

int t1 = toll[tim[i][j].hour]* tim[i][j].mina;
for(int k = 0; k < tim[i][j].hour; ++k) t1 += toll[k]*60;

int t2 = toll[tim[i][j+1].hour]* tim[i][j+1].mina;
for(int k = 0; k < tim[i][j+1].hour; ++k) t2 += toll[k]*60;
if(tim[i][j].day == tim[i][j+1].day) {
a1 = t2-t1;
}else {
a1 = (allhour*60 - t1) + t2;
a1 += (tim[i][j+1].day-tim[i][j].day-1) * allhour*60;
}

t1 = tim[i][j].mina;
for(int k = 0; k < tim[i][j].hour; ++k) t1 += 60;

t2 = tim[i][j+1].mina;
for(int k = 0; k < tim[i][j+1].hour; ++k) t2 += 60;
if(tim[i][j].day == tim[i][j+1].day) {
a2 = t2-t1;
}else {
a2 = (24*60 - t1) + t2;
a2 += (tim[i][j+1].day-tim[i][j].day-1) * 24*60;
}

ans[i][tim[i][j].mon].push_back( MP( MP(tim[i][j], tim[i][j+1]), MP(a2,a1) ) );
}else j--;
}
}

for(it = mp.begin(); it != mp.end(); ++it) {
int id = it->second;
for(int j = 1; j <= 12; ++j) {

if(ans[id][j].size() > 0) {
int all = 0;
for(int k = 0; k < it->first.length(); ++k) printf("%c", it->first[k]);
printf(" %02d\n", j);

for(int k = 0; k < ans[id][j].size(); ++k) {
Node t1 = ans[id][j][k].first.first; Node t2 = ans[id][j][k].first.second;
int t3 = ans[id][j][k].second.first; int t4 = ans[id][j][k].second.second;
all += t4;
printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n",
t1.day,t1.hour,t1.mina, t2.day,t2.hour,t2.mina ,t3, t4/100.0);
}
printf("Total amount: $%.2f\n", all/100.0);
}
}
}
}

1017数据无话可说，我对于时间的界限还是不好掌握

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MP(x, y) make_pair(x, y)
#define FI first
#define SE second

const int N = 1e4+5;
const int INF = 0x3f3f3f3f;

struct Node{
int tim, spend, id;
Node(int a=0, int b=0, int c=0):tim(a), spend(b), id(c){}
}E
;
int comtime
;
int ans
;
int cmp(Node a, Node b) {
return a.tim < b.tim;
}
int windows[105];

int main() {
int n, k;
while(~scanf("%d %d", &n, &k)) {
for(int i = 1; i <= n; ++i) {
int a, b, c, d;
scanf("%02d:%02d:%02d %d", &a,&b,&c,&d);
int tt = a*60*60 + b*60 + c;
if(d > 60) d = 60;
E[i] = Node(tt, d, i);
comtime[i] = tt;
}
sort(E+1, E+n+1, cmp);

for(int i = 1; i <= k; ++i) windows[i] = 8*3600;

int ntime = 0;
for(int i = 1; i <= n; ++i) {
//     printf("%d ", E[i].tim);
int mx = INF, po;
for(int j = 1; j <= k; ++j) {
if(mx > windows[j]) {
mx = windows[j]; po = j;
}
}
mx = max(E[i].tim, mx);
ans[E[i].id] = mx;
//   printf("%d\n", mx);

windows[po] = mx + E[i].spend*60;
}

int all = 0;
int cnt = 0;
for(int i = 1; i <= n; ++i) {
//       printf("%d %d\n", ans[i], comtime[i]);

if(comtime[i] > 17*3600) {
continue;
}
cnt ++;
int tt = ans[i]-comtime[i];
all += tt;
}

if(cnt) printf("%.1f\n", all/60.0/cnt);
else printf("0.0\n");
}
return 0;
}

1018把我写成狗，我真的得掌握dijsktra+dfs这种方法，比我这个好多了

当然别忘了send，back可能同时存在，这很重要

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MP(x, y) make_pair(x, y)
#define FI first
#define SE second

const int N = 505;
const int INF = 0x3f3f3f3f;

int Cmax;
int n, sp, m;
int C
;
struct Node{
int to, nx, dis;
}E[N*N*2];
int head
, tot;
void add(int a, int b, int c) {
E[tot].to = b; E[tot].nx = head[a]; E[tot].dis = c;
head[a] = tot++;
}
struct HeapNode{
int po, di;
HeapNode(int a=0, int b=0):po(a), di(b){}
bool operator <(const HeapNode & T) const{
return di > T.di;
}
};
int dis
; int vis
;
struct Pode{
int send, bak, pre, preid;
Pode(int a=0, int b=0, int c=0, int d=0):send(a), bak(b), pre(c), preid(d){}
};
vector<Pode> bike
;
void dijstra(int st) {
memset(dis, INF, sizeof(dis));
memset(vis, 0, sizeof(vis));

for(int i = 0; i <= n; ++i) bike[i].clear();
dis[st] = 0; bike[st].push_back(Pode(0,0,0,0));

priority_queue<HeapNode> Q;
Q.push(HeapNode(st, dis[st]));
while(!Q.empty()) {
int x = Q.top().po; Q.pop();
if(vis[x]) continue;
vis[x] = 1;

for(int i = head[x]; ~i; i = E[i].nx) {
int to = E[i].to;

if(dis[to] > dis[x]+E[i].dis) {
dis[to] = dis[x]+E[i].dis;

bike[to].clear();
for(int j = 0; j < bike[x].size(); ++j) {
int t1 = bike[x][j].send; int t2 = bike[x][j].bak;

if(Cmax <= C[to]) t2 += C[to]-Cmax;
else {
int t3 = Cmax-C[to]; int t4 = min(t2, t3);
t2 -= t4; t3 -= t4;
if(t3) t1 += t3;

}

bike[to].push_back(Pode(t1, t2, x, j));

}
Q.push(HeapNode(to, dis[to]));
}
else if(dis[to] == dis[x]+E[i].dis) {
map<pair<int,int> , int> mmp; mmp.clear();
for(int j = 0; j < bike[to].size(); ++j) {
mmp[ MP(bike[to][j].send, bike[to][j].bak) ] ++;
}
for(int j = 0; j < bike[x].size(); ++j) {
int t1 = bike[x][j].send; int t2 = bike[x][j].bak;

if(Cmax <= C[to]) t2 += C[to]-Cmax;
else {
int t3 = Cmax-C[to]; int t4 = min(t2, t3);
t2 -= t4; t3 -= t4;
if(t3) t1 += t3;
}

if(mmp.find(MP(t1, t2)) == mmp.end())
bike[to].push_back(Pode(t1, t2, x, j));
}
}
}
}
}
void dfs(int x, int id) {
//  printf("%d %d\n", x, id);
if(x == 0) {
printf("0"); return;
}

int pre = bike[x][id].pre; int preid = bike[x][id].preid;
dfs(pre, preid);

printf("->%d", x);
}
void solve() {
int a1 = INF, a2 = INF; int id;
for(int i = 0; i < bike[sp].size(); ++i) {
if(a1 > bike[sp][i].send) {
a1 = bike[sp][i].send; a2 = bike[sp][i].bak; id = i;
}else if(a2 >bike[sp][i].bak && a1 == bike[sp][i].send) {
a2 = bike[sp][i].bak; id = i;
}
}

printf("%d ", bike[sp][id].send);
dfs(sp, id);
printf(" %d\n", bike[sp][id].bak);
//  if(bike[sp][id].send && bike[sp][id].bak) while(1);
}

int main() {
C[0] = 0;
while(~scanf("%d %d %d %d", &Cmax, &n, &sp, &m)) {
Cmax /= 2;
memset(head, -1, sizeof(head)); tot = 0;
for(int i = 1; i <= n; ++i) {
scanf("%d", &C[i]);
}
for(int i = 0; i < m; ++i) {
int a, b, c; scanf("%d %d %d",&a,&b,&c);
add(a, b, c); add(b, a, c);
}

dijstra(0);
solve();
}
return 0;
}

1019

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e6+5;
const int M = 3e5+5;

int main() {
int n, b;
while(~scanf("%d %d",&n,&b)) {
int tmp = n;
vector<int> ans;
while(tmp) {
ans.push_back(tmp%b);
tmp /= b;
}

int fl = 1;
int len = ans.size();
for(int i = 0; i < len/2; ++i) {
if(ans[i] != ans[len-i-1]) {
fl = 0; break;
}
}

if(fl) printf("Yes\n");
else printf("No\n");

if(n == 0) {
printf("0\n");
continue;
}
for(int i = len-1; i >= 0; --i) {
if(i != len-1) printf(" ");
printf("%d", ans[i]);
}
printf("\n");
}
return 0;
}

1020

#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;

int a[35];
int b[35];

struct Node{
int l,r,L,R;
Node(int a=0, int b=0, int c=0, int d=0):l(a), r(b), L(c), R(d){}
};
int main() {
int n;
while(~scanf("%d", &n)) {
//    memset(head, -1, sizeof(head)); tot = 0;

for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
for(int i = 1; i <= n; ++i) scanf("%d", &b[i]);

queue<Node> Q;
Q.push(Node(1, n, 1, n));
int fl = 1;
int cnt = 0;
while(!Q.empty()) {
Node t = Q.front(); Q.pop();
if(t.l > t.r) continue;
/*  printf("%d\n", cnt);
if(cnt < 10) {
printf("%d\n", t.r);
cnt ++;
}*/
int tag = a[t.r];

if(fl) {
fl = 0;
}else printf(" ");

printf("%d", tag);

if(t.l == t.r) continue;

int M;
for(int i = t.L; i <= t.R; ++i) {
if(b[i] == tag) {
M = i-1; break;
}
}

int m = t.l+(M-t.L);

Q.push(Node(t.l, m, t.L, M));
Q.push(Node(m+1, t.r-1, M+2, t.R));
}
printf("\n");
}
return 0;
}