Codeforces Round #583 (Div. 1 + Div. 2, based on Olympiad of Metropolises)
2024-05-12 03:06:30
枚举题意，五十分钟看懂题，十分钟过三道，tarjan还写挂了，大型翻车现场。
A、你有n卢布，要把卢布换成美元和欧元，一美元和一欧元对应的卢布数是给定的
美元有几种面值：1、2、5、10、20、50、100
欧元有几种面值：5、10、20、50、100、200
你可以兑换任意数量的欧元和美元，但你兑换的美元和欧元必须可以用上面的纸币表示出来，问你最后手里最少剩下多少卢布
美元只要能换就换，因为1美元可以表示任何数字，欧元只能5欧5欧地换，枚举一下换成美元的张数即可
代码：
#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}int n,d,e;
int ans=inf;signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>n>>d>>e;int k=0;while(1){if(k*d>n) break;int tmp=n;tmp-=k*d;int ou=tmp/e;ou-=ou%5;tmp-=ou*e;ans=min(ans,tmp);k++;}cout<<ans;return 0;
}B、b个男孩和g个女孩吃饭，一共来n个人，现在有n+1张桌子，第i张桌子可以坐i个男生和n-i个女生
问你最少需要多少张桌子才能使得不管来几个男生女生都能坐
SB题意，看半天没搞懂，两边扫一下就完事了，分别表示男生全来和女生全来的情况
代码：
#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}int b,g,n;
int bb[maxn],gg[maxn];signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>b>>g>>n;re(i,0,n) bb[i]=i,gg[i]=n-i;int x=min(n,b);int y=n-x;int l;re(i,0,n) if(bb[i]==x&&gg[i]==y) {l=i;break;}y=min(n,g);x=n-y;int r;re(i,0,n) if(bb[i]==x&&gg[i]==y) {r=i;break;}cout<<abs(r-l)+1;return 0;
}C、给你一堆括号序列，问你能否通过交换一堆括号使得其编程有序，也可以不交换
上个栈，最后栈里只剩下一个左括号并且过程中有一个右括号未匹配就YES，直接匹配也输出YES，否则NO
代码：
#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}int n;
string s;
stack<char> st;
int l=0;signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>n>>s;fo(i,0,n){if(s[i]=='('){st.push('(');}else{if(st.empty()) l++;else st.pop();}}if((l==0&&st.empty())||(l==1&&st.size()==1))cout<<"Yes"<<endl;else cout<<"No"<<endl;return 0;
}D、看到standing里面很多写dp的，完全看不懂……
一开始大爆搜，搜出所有可能路径，路径上每个点加一，如果有一个点每条路径都经过，那么输出1
比赛时rhy糊了个算法，把所有可以到达终点的路径打上标记，对这些标记建立四联通的无向图，跑一遍tarjan
不能到达答案为0，如果有割点，答案为1，否则为2
tarjan要注意一个问题，起点不能做为割点
打标记也要注意一个问题，只有抵达终点的路径才打标记，否则会错误判断割点。
这种样例就会出锅：
4 4
....
....
.#..
.#..
左下角虽然有个割点，但是完全起不到作用。此外dfs的时候还得注意或运算短路的性质，免得搜完向下的路径就不搜向右的路径了。
代码：
#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=1000005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}int n,m,timer=0,cut=0;
bool can=0;
char rhy[maxn];
vector<char> s[maxn];
vei dfn[maxn],low[maxn];
int a[4]={1,0,-1,0};
int b[4]={0,1,0,-1};bool ok(int i,int j){if(i<0||i>=n||j<0||j>=m) return 0;if(s[i][j]=='#') return 0;return 1;
}bool dfs(int i,int j){if(i<0||i>=n||j<0||j>=m) return 0;if(s[i][j]=='#') return 0;if(s[i][j]=='0') return 1;if(i==n-1&&j==m-1){can=1;s[i][j]='0';return 1;}bool now=0;now|=dfs(i+1,j);now|=dfs(i,j+1);if(now){
// cout<<i<<' '<<j<<endl;s[i][j]='0';if(ok(i+1,j))s[i+1][j]='0';if(ok(i,j+1))s[i][j+1]='0';return 1;}else{s[i][j]='#';return 0;}
}void tarjan(int x,int y){dfn[x][y]=low[x][y]=++timer;fo(i,0,4){int dx=x+a[i],dy=y+b[i];if(dx<0||dx>=n||dy<0||dy>=m||s[dx][dy]!='0')continue;if(!dfn[dx][dy]){tarjan(dx,dy);low[x][y]=min(low[x][y],low[dx][dy]);if(low[dx][dy]>=dfn[x][y]&&(x!=0||y!=0)){
// cout<<x<<' '<<y<<endl;cut++;}}else{low[x][y]=min(low[x][y],dfn[dx][dy]);}}
}signed main(){scc(n,m);fo(i,0,n){scanf("%s",rhy);fo(j,0,m)dfn[i].pub(0),low[i].pub(0),s[i].pub(rhy[j]);}dfs(0,0);s[0][0]=s[n-1][m-1]='0';
// fo(i,0,n) fo(j,0,m) cout<<s[i][j]<<(j==m-1?'\n':' ');if(!can) cout<<0;else{tarjan(0,0);if(!cut) cout<<2;else cout<<1;}return 0;
}
/*
2 4
....
..#.3 3
.#.
#..
...4 4
....
....
.#..
.#..
*/转载于:https://www.cnblogs.com/oneman233/p/11465061.html
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