不存在的数
- 签到题,模拟
- 对输入的数进行标记,从 1 遍历到 N,输出没有被标记的数字即可
std标程:
代码语言:javascript复制#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <sstream>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
using namespace std;
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define re register
#define fi first
#define se second
#define endl 'n'
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
const int N = 1e6 3;
const int INF = 0x3f3f3f3f, mod = 1e9 7;
const double eps = 1e-6, PI = acos(-1);
int a[N];
void solve(){
int n, m; cin >> n >> m;
for (int i = 0; i < m; i ) {
int x; cin >> x; a[x] = 1;
}
for (int i = 1; i <= n; i ) {
if (a[i] == 0) cout << i << ' ';
}
}
int main(){
IOS;
int _ = 1;
// cin >> _;
while(_ --){
solve();
}
return 0;
}
一血代码:
代码语言:javascript复制#include<bits/stdc .h>
using namespace std;
int main()
{
int n, m;
cin >> n >> m;
map<int, int>mp;
for( int i = 1 ; i <= m ; i ) {
int x;
cin >> x;
mp[x] ;
}
for( int i = 1 ; i <= n ; i ) {
if( mp[i] ) continue;
else cout << i << ' ';
}
}
疯狂校园跑
配速为1km所用时间,所以配速为时间除以路程;
知道这一点之后就可以写代码了。
std标程:
代码语言:javascript复制#include <stdio.h>
int main()
{
int t; scanf("%d", &t);
while (t --)
{
int s, t; scanf("%d %d", &s, &t);
double sps = t * 1.0 / s; //多少秒每米
double spm = sps * 1000; //多少秒每公里
printf("%d:d:dn", (int)(spm / 60), (int)(spm - (int)(spm / 60) * 60), ((int)(spm * 1000) % 1000));
}
return 0;
}
一血代码:
代码语言:javascript复制#include <bits/stdc .h>
#define int long long
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define endl 'n'
using namespace std;
void solve(){
int n, s;
cin >> n >> s;
double res = 1.0 * 1000 * s / n;
int h = res / 60;
int m = res - h * 60;
int ms = 1.0 * (res - h * 60 - m) * 1000;
printf("%d:d:dn", h, m, ms);
}
signed main(){
IOS;
int t = 1;
cin >> t;
while(t --) solve();
return 0;
}
输出语句
- 模拟
- 考察转义字符,这里只考察了大家单引号
'
,双引号"
,回车符n
以及反斜杠\
,对输出加以判定即可
std标程:
代码语言:javascript复制#include <bits/stdc .h>
using namespace std;
void solve()
{
string s;
getline(cin, s);
int l = 8, r = s.size() - 4;
for (int i = l; i <= r; i )
{
if (i != r && s[i] == '\')
{
if (s[i 1] == ''')
{
cout << ''';
i ;
}
else if (s[i 1] == '"')
{
cout << '"';
i ;
}
else if (s[i 1] == '\')
{
cout << '\';
i ;
}
else if (s[i 1] == 'n')
{
cout << 'n';
i ;
}
}
else
{
cout << s[i];
}
}
}
int main()
{
solve();
return 0;
}
一血代码:
代码语言:javascript复制#include<iostream>
#include<string>
using namespace std;
int main(void)
{
string a;
getline(cin,a);
int flag=0;
for(int i=0;i<a.size();i )
{
if(a[i-1]=='f'&&a[i]=='('&&a[i 1]=='"')
{
for(int j=i 2;j<a.size();j )
{
if(a[j]=='"'&&a[j 1]==')'&&a[j 2]==';')
{
break;
}
if(a[j]=='\')
{
j ;
if(a[j]=='0')
{
cout<<" ";
}
else if(a[j]=='n')
{
cout<<endl;
}
else
{
cout<<a[j];
}
}
else cout<<a[j];
}
}
}
return 0;
}
疯狂汽车运输队
本题为简单的01背包,不过是需要算一下超重多少,但是最多超重100%,那么代码就如下。
std标程:
代码语言:javascript复制#include<bits/stdc .h>
using namespace std;
int f[10000];
int main()
{
int T;
cin>>T;
while(T--)
{
int n,m;
cin>>n>>m;
for(int i=1;i<=m*2;i ) f[i]=0;
for(int i=1;i<=n;i )
{
int x,y;
cin>>x>>y;
for(int j=m*2;j>=x;j--) f[j]=max(f[j],f[j-x] y);
}
int ans=0;
for(int i=m;i<=2*m;i )
{
int x=(i-m)*1.0/(m*1.0)*10;
// cout<<x<<" "<<f[i]<<" ";
x=f[i]-(int)(f[i]*x*0.1);
ans=max(x,ans);
// cout<<x<<"n";
}
cout<<ans<<"n";
}
return 0;
}
为什么演奏春日影
- 签到题,trick
- 选择 k = 1,则后面每次都开摆,任何正整数取模结果都是 0,因此结果全为 Yes
std标程:
代码语言:javascript复制#include <bits/stdc .h>
using namespace std;
int main()
{
int n, m; cin >> n >> m;
for (int i = 0; i < n; i ) {
int x; cin >> x;
}
cout << "Yes";
return 0;
}
一血代码:
代码语言:javascript复制#include<bits/stdc .h>
using namespace std;
map<int, int>mp;
void calc( int x )
{
for( int i = 2 ; i <= sqrt(x) ; i ) {
if( x % i == 0 ) {
mp[i] ;
mp[x % i] ;
}
}
}
int main()
{
int n, m;
cin >> n >> m;
vector<int> a(n);
for( int i = 0 ; i < n ; i ) {
cin >> a[i];
}
for( int i = 0 ; i < n ; i ) {
calc(a[i]);
}
int mx = 0;
for( auto i : mp ) {
mx = max(mx, i.second);
}
if( mx m < n ) cout << "No";
else cout << "Yes";
}
BinGo游戏
- 签到题,模拟
- 根据题意模拟循环判断即可
std标程:
代码语言:javascript复制#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <sstream>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
using namespace std;
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define re register
#define fi first
#define se second
#define endl 'n'
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
const int N = 1e6 3;
const int INF = 0x3f3f3f3f, mod = 1e9 7;
const double eps = 1e-6, PI = acos(-1);
int a[10][10];
bool check1() {
for (int i = 0; i < 5; i ) {
bool flag = 1;
for (int j = 0; j < 5; j ) {
if (a[i][j] % 2 != 0) {
flag = 0; break;
}
}
if (flag) return 1;
}
return 0;
}
bool check2() {
for (int i = 0; i < 5; i ) {
bool flag = 1;
for (int j = 0; j < 5; j ) {
if (a[j][i] % 2 == 0) {
flag = 0; break;
}
}
if (flag) return 1;
}
return 0;
}
bool check3() {
bool mainDiag = a[0][0] % 2 == 0;
bool offDiag = a[0][4] % 2 == 0;
for (int i = 1; i < 5; i ) {
if ((a[i][i] % 2 == 0) == mainDiag || (a[i][4-i] % 2 == 0) == offDiag) return false;
mainDiag = !mainDiag;
offDiag = !offDiag;
}
return true;
}
void solve(){
for (int i = 0; i < 5; i ) {
for (int j = 0; j < 5; j ) cin >> a[i][j];
}
if (check1() || check2() || check3()) {
cout << "BINGO" << endl;
} else {
cout << "OHNO" << endl;
}
}
int main(){
IOS;
int _ = 1;
// cin >> _;
while(_ --){
solve();
}
return 0;
}
一血代码:
代码语言:javascript复制#include <bits/stdc .h>
#define int long long
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define endl 'n'
using namespace std;
int a[10][10];
bool ok;
bool check(int x){
int f = 0;
for(int i = 0; i < 5; i ){
if(a[i][x] % 2 != 1) break;
if(i == 4) f = 1;
}
for(int i = 0; i < 5; i ){
if(a[x][i] % 2 != 0) break;
if(i == 4) f = 1;
}
if(f) return true;
return false;
}
void solve(){
for(int i = 0; i < 5; i )
for(int j = 0; j < 5; j )
cin >> a[i][j];
for(int i = 0; i < 5; i ){
if(check(i)){
ok = true;
// cout << i << endl;
}
}
// if(ok) cout << 1 << endl;
bool ok1 = false, ok2 = false;
for(int i = 1; i < 5; i ){
if(a[i][i] % 2 == a[i - 1][i - 1] % 2){
break;
}
if(i == 4) ok1 = true;
}
for(int i = 1; i < 5; i ){
if(a[i][4 - i] % 2 == a[i - 1][5 - i] % 2){
break;
}
if(i == 4) ok2 = true;
}
if(ok1 && ok2) ok = true;
if(ok) cout << "BINGO" << endl;
else cout << "OHNO" << endl;
}
signed main(){
IOS;
int t = 1;
// cin >> t;
while(t --) solve();
return 0;
}
下载序列
- 思维,小根堆
- 每个文件总共需要的时间(包含下载和文件验证)可以预处理出来,那么我们可以维护一个容量为 m 的队列,每次让当前队列里时间最小的文件出列,并将时间累加到新的入队时间上,那么最终最后一个出队的文件将是时间最大的,记为 tmax,将 tmax * 维护费k 输出即可。
std标程:
代码语言:javascript复制#include <bits/stdc .h>
using namespace std;
typedef long long LL;
const int N = 1e5 10;
int x, m, k, n, e[N];
priority_queue<LL, vector<LL>, greater<LL>> q;
int main() {
cin >> x >> m >> k;
cin >> n;
for (int i = 1; i <= n; i )
{
int a, t;
cin >> a >> t;
e[i] = (a x - 1) / x t;
}
for (int i = 1; i <= m; i )
{
q.push(e[i]);
}
LL ans = 0;
for (int i = m 1; i <= n; i )
{
ans = q.top(); q.pop();
q.push(e[i] ans);
}
while (q.size())
{
ans = q.top(); q.pop();
}
cout << ans * k;
return 0;
}
会 赢 的
我们只需要以下操作即可:
- 选取数组中的 100(至多一个)
- 选取 1 ~ 9 之间的一个数
- 选取 10 ~ 90 之间能被10整除的数
- 如果 2、3 操作选不出任何一个数,那么可以在剩下的数里选任意一个(没有则不选)
std标程:
代码语言:javascript复制#include <stdio.h>
int n, a[1010];
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i )
scanf("%d", &a[i]);
for (int i = 1; i < n; i )
for (int j = i 1; j <= n; j )
if (a[i] > a[j])
{
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
int ans = 0, p[5] = {0};
for (int i = 1; i <= n; i )
{
if (p[1] == 0 && a[i] >= 1 && a[i] <= 9)
p[1] , ans ;
else if (p[2] == 0 && a[i] >= 10 && a[i] <= 99 && a[i] % 10 == 0)
p[2] , ans ;
else if (p[3] == 0 && a[i] == 100)
p[3] , ans ;
else
p[0] ;
}
if (p[1] == 0 && p[2] == 0 && p[0] > 0) ans ;
printf("%d", ans);
return 0;
}
一血代码:
代码语言:javascript复制#include<bits/stdc .h>
using namespace std;
typedef long long LL;
const int N=110;
int a[5];
int n,res=0,cnt=0;
int main(){
cin>>n;
for(int i=1;i<=n;i ){
int x;
cin>>x;
if(x==0)a[0] ;
else if(x==100)a[1]=1;
else if(x>0&&x<10)a[2] ;
else a[3] ;
if(x==0)cnt ;
}
if(a[0])res =a[0];
if(a[1])res ;
if(a[2]&&a[3]){
if(cnt)res =2;
else res =1;
}else if(a[2]||a[3])res =1;
cout<<res<<endl;
return 0;
}