一開始幾道題寫到吐血，真的自己現在好弱
1011 水題不說了

#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 比較麻煩，一開始處理一下

#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[N], tot;void add(int fr,int to) {E[tot].to = to; E[tot].nx = head[fr]; head[fr] = tot++; } int ans[N]; int cnt; int nop; int vis[N]; 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 這道題我的理解完全不一樣，對于5點下班這件事，我真的無語，這種坑就需要猜了

#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翻轉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 很煩一題，我的世界一開始計算錯了

#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數據無話可說，我對于時間的界限還是不好掌握

#include<bits/stdc++.h> using namespace std; typedef long long ll; #define MP(x, y) make_pair(x, y) #define FI first #define SE secondconst 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[N]; int comtime[N]; int ans[N]; 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把我寫成狗，我真的得掌握dijsktra+dfs這種方法，比我這個好多了
當然別忘了send，back可能同時存在，這很重要

#include<bits/stdc++.h> using namespace std; typedef long long ll; #define MP(x, y) make_pair(x, y) #define FI first #define SE secondconst int N = 505; const int INF = 0x3f3f3f3f;int Cmax; int n, sp, m; int C[N]; struct Node{int to, nx, dis; }E[N*N*2]; int head[N], 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[N]; int vis[N]; 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[N]; 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; }

轉載于:https://www.cnblogs.com/Basasuya/p/8433712.html

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