当前位置:网站首页>tese_Time_2h
tese_Time_2h
2022-07-01 05:25:00 【爱思考的小伙】
文章目录
Relic Discovery HDU - 5982
#include<iostream>
#include<cstdio>
using namespace std;
int main(){
int t; scanf("%d",&t);
while(t--){
int n,a,b,ans=0; scanf("%d",&n);
for(int i=1; i<=n; i++){
scanf("%d%d",&a,&b);
ans += a*b;
}
printf("%d\n",ans);
}
return 0;
}
Text Reverse HDU - 1062
- 下面给出另一个超时程序,但是我没发现问题,大家可以找找
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
using namespace std;
const int N=1e6+10, INF=0x3f3f3f3f;
char a[N];
int main() {
// freopen("data.in", "r", stdin);
int n; scanf("%d\n", &n);
for(int i=1; i<=n; i++) {
stack<char> sta;
while(1) {
char ch=getchar();
if(ch==' ') {
while(!sta.empty()) {
printf("%c", sta.top()); sta.pop();
}
printf(" ");
} else if(ch=='\n') {
while(!sta.empty()) {
printf("%c", sta.top()); sta.pop();
}
printf("\n"); break;
} else sta.push(ch);
}
}
return 0;
}
int main_Time_LE() {
// freopen("data.in", "r", stdin);
int n;scanf("%d\n", &n);
for(int i=1; i<=n; i++) {
while(1){
memset(a, 0, sizeof(a));
scanf("%s", a);
int len = strlen(a);
reverse(a, a+len);
char ch=getchar();
if(ch=='\n') break;
printf("%s ", a);
}
puts(a);
}
return 0;
}
Download Manager HDU - 3233
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
const int N=1e6+10, INF=0x3f3f3f3f;
int main(){
// freopen("data.in", "r", stdin);
int t,n,b,cnt=0;
while(~scanf("%d%d%d", &t,&n,&b) && t&&n&&b){
double s,p, sum=0;
for(int i=1; i<=t; i++){
scanf("%lf%lf", &s, &p);
sum += s*(100-p)*0.01;
}
double ans=sum/b;
printf("Case %d: %.2lf\n\n",++cnt,ans);
}
return 0;
}
看病要排队 HDU - 1873
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
using namespace std;
const int N=2e4+10, INF=0x3f3f3f3f;
struct T{
int no, id;
T(){
}
T(int a,int b):no(a), id(b){
}
bool operator< (const T& t) const{
if(no!=t.no) return no < t.no;
return id > t.id;
}
};
int main() {
// freopen("data.in", "r", stdin);
string op; int n, a,b,cnt=0;
while(cin>>n){
priority_queue<T> que[4];
cnt=0;
for(int i=1; i<=n; i++){
cin>>op;
if(op=="IN"){
cin>>a>>b;
que[a].push(T(b, ++cnt));
}else if(op=="OUT"){
cin>>a;
if(que[a].empty()) {
cout<<"EMPTY"<<endl;
}else{
cout<<que[a].top().id<<endl; que[a].pop();
}
}
}
}
return 0;
}
Red and Black HDU - 1312
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
using namespace std;
const int N=110, INF=0x3f3f3f3f;
int n,m,dis[][2]={
-1,0, 0,-1, 1,0, 0,1}, vis[N][N];
char a[N][N];
void pr(){
for(int i=0; i<m; i++) {
for(int j=0; j<n; j++) printf("%d ", vis[i][j]);
puts("");
}
}
int bfs(int x,int y){
int ans=1;
memset(vis, 0x00, sizeof(vis));
queue<pair<int,int> > que; que.push(make_pair(x,y)); vis[x][y]=1;
while(!que.empty()){
pair<int,int> p=que.front(); que.pop();
for(int i=0; i<4; i++){
int tx = p.first+dis[i][0];
int ty = p.second+dis[i][1];
if(a[tx][ty]!='.' || vis[tx][ty] || tx<0||tx>=m || ty<0||ty>=n) continue;
ans++, vis[tx][ty]=vis[p.first][p.second]+1;
que.push(make_pair(tx, ty));
}
}
return ans;
}
int main() {
// freopen("1.cpp", "r", stdin);
while(~scanf("%d%d", &n,&m) && n && m){
for(int i=0; i<m; i++) scanf("%s", a[i]);
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
if(a[i][j]=='@') {
printf("%d\n", bfs(i,j)); break;
}
}
}
// pr();
}
return 0;
}
最短路 HDU - 2544
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
using namespace std;
const int N=1e5+10, INF=0x3f3f3f3f;
int head[N],cnt,dis[N],vis[N],n,m;
struct T{
int to,w,next;
T(){
}
T(int a,int b):to(a), w(b){
}
bool operator<(const T &t) const{
return w > t.w;
}
}G[N];
void add(int u,int v,int w){
G[++cnt].to = v;
G[cnt].w = w;
G[cnt].next = head[u];
head[u] = cnt;
}
void pri_dijkstra(int s){
memset(vis, 0x00, sizeof(vis));
memset(dis, 0x3f, sizeof(dis));
dis[s]=0;
priority_queue<T> que; que.push(T(s, dis[s]));
while(!que.empty()){
int u=que.top().to; que.pop();
if(vis[u]) continue;
vis[u]=1;
for(int i=head[u]; ~i; i=G[i].next){
int v=G[i].to, w=G[i].w;
if(dis[v] > dis[u]+w){
dis[v] = dis[u]+w;
que.push(T(v, dis[v]));
}
}
}
}
int main() {
// freopen("1.cpp", "r", stdin);
int a,b,c;
while(~scanf("%d%d", &n, &m) && n && m){
memset(head, -1, sizeof(head));
for(int i=1; i<=m; i++){
scanf("%d%d%d", &a,&b,&c);
add(a, b, c), add(b, a, c);
}
pri_dijkstra(1);
printf("%d\n", dis[n]);
}
return 0;
}
边栏推荐
- Global and Chinese market of solder wire 2022-2028: Research Report on technology, participants, trends, market size and share
- tar命令
- QT waiting box production
- Unity drags and modifies scene camera parameters under the editor
- C# wpf 使用DockPanel实现截屏框
- 轻松上手Fluentd,结合 Rainbond 插件市场,日志收集更快捷
- HDU - 1024 Max Sum Plus Plus(DP)
- [RootersCTF2019]babyWeb
- Detailed explanation of set
- Simple implementation of database connection pool
猜你喜欢

Series of improving enterprise product delivery efficiency (1) -- one click installation and upgrade of enterprise applications

导电滑环短路的原因以及应对措施

eBPF Cilium实战(2) - 底层网络可观测性

CockroachDB 分布式事务源码分析之 TxnCoordSender

Application of industrial conductive slip ring

Set集合詳細講解

LevelDB源码分析之memtable

Mathematical knowledge: finding the number of divisors

Lock free concurrency of JUC (leguan lock)

数字金额加逗号;js给数字加三位一逗号间隔的两种方法;js数据格式化
随机推荐
[RootersCTF2019]babyWeb
Day 05 - file operation function
[Yugong series] February 2022 Net architecture class 005 ABP vNext Net core web application getting started configuration
C WPF uses dockpanel to realize screenshot box
如何开始学剪辑?零基础详细解析
Understand several related problems in JVM - JVM memory layout, class loading mechanism, garbage collection
Vérification simple de la lecture et de l'écriture de qdatastream
0xc000007b应用程序无法正常启动解决方案(亲测有效)
tar命令
Global and Chinese market of metal oxide semiconductor field effect transistors 2022-2028: Research Report on technology, participants, trends, market size and share
3D建模與處理軟件簡介 劉利剛 中國科技大學
HDU - 1024 Max Sum Plus Plus(DP)
How to traverse massive data in redis
Rainbow combines neuvector to practice container safety management
Worried about infringement? Must share copyrightless materials on the website. Don't worry about the lack of materials for video clips
Global and Chinese market of paper machine systems 2022-2028: Research Report on technology, participants, trends, market size and share
What things you didn't understand when you were a child and didn't understand until you grew up?
2/15 (awk, awk conditions, awk processing design can perform additional tasks, and use awk array +for loop to realize advanced search)
Simple read / write verification of qdatastream
云原生存储解决方案Rook-Ceph与Rainbond结合的实践