输出a , b , c , d的最小值即可。
#include "iostream"
#include "cstring"
#include "string"
#include "vector"
#include "cmath"
#include "algorithm"
#include "map"
#include "set"
#include "queue"
#include "stack"
#define fi first
#define se second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define seteps(N) setprecision(N)
#define lson (ind<<1)
#define rson (ind<<1|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 lll;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int N=1010;
const int M=1<<19;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll oone=1;
const double eps=1e-6;
const double pi=acos(-1);
void solve(){
int a,b,c,d;
cin>>a>>b>>c>>d;
int res=min(a,b);
res=min(res,c);
res=min(res,d);
cout<<res<<endl;
}
int main(){
IOS;
//freopen("try.in", "r", stdin);
//freopen("try2.out", "w", stdout);
//int t;cin>>t;
//while(t--)
solve();
return 0;
}
根据题意模拟过程。需要注意的一点是,在咖啡店充电时,电量不能超过充电宝的上限。
#include "iostream"
#include "cstring"
#include "string"
#include "vector"
#include "cmath"
#include "algorithm"
#include "map"
#include "set"
#include "queue"
#include "stack"
#define fi first
#define se second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define seteps(N) setprecision(N)
#define lson (ind<<1)
#define rson (ind<<1|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 lll;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int N=1010;
const int M=1<<19;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll oone=1;
const double eps=1e-6;
const double pi=acos(-1);
void solve(){
ll n,m,t;
cin>>n>>m>>t;
ll cur=n,last=0;
bool ok=true;
rep(i,0,m){
ll a,b;
cin>>a>>b;
cur-=(a-last);
if(cur<=0) ok=false;
cur+=(b-a);
if(cur>n) cur=n;
last=b;
}
if(cur) cur-=(t-last);
if(cur<=0) ok=false;
cout<<(ok ? "Yes" :"No")<<endl;
}
int main(){
IOS;
//freopen("try.in", "r", stdin);
//freopen("try2.out", "w", stdout);
//int t;cin>>t;
//while(t--)
solve();
return 0;
}
直接算C ( l − 1 , l − 12 )即可。由于题目中没有模数,偷懒使用了JAVA的大整数
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int l = cin.nextInt();
int u=l-1,d=l-12;
BigInteger big = new BigInteger("1");
for(int i=u;i>=u-d+1;i--){
big = big.multiply(new BigInteger(String.valueOf(i)));
}
for (int i = 1; i <= d; i++) big = big.divide(new BigInteger(String.valueOf(i)));
System.out.println(big.toString());
}
}
找到两个蓝色邮票之间最少的白色邮票个数,然后从前到后模拟,算下个数即可。
#include "iostream"
#include "cstring"
#include "string"
#include "vector"
#include "cmath"
#include "algorithm"
#include "map"
#include "set"
#include "queue"
#include "stack"
#define fi first
#define se second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define seteps(N) setprecision(N)
#define lson (ind<<1)
#define rson (ind<<1|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 lll;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int N=3*1e5+10;
const int M=1<<19;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll oone=1;
const double eps=1e-6;
const double pi=acos(-1);
int a[N];
void solve(){
int n,m;
cin>>n>>m;
rep(i,1,m+1) cin>>a[i];
if(!m){
cout<<1<<endl;
return;
}
if(n==m){
cout<<0<<endl;
return;
}
sort(a+1,a+m+1);
int mn=INF;
if(a[1]!=1) mn=min(mn,a[1]-1);
if(a[m]!=n) mn=min(mn,n-a[m]);
rep(i,2,m+1){
if(a[i]-a[i-1]-1==0) continue;
mn=min(mn,a[i]-a[i-1]-1);
}
int ans=0;
rep(i,2,m+1){
int gap=a[i]-a[i-1]-1;
ans+=gap/mn+(gap%mn!=0);
}
ans+=(n-a[m])/mn+((n-a[m])%mn!=0);
ans+=(a[1]-1)/mn+((a[1]-1)%mn!=0);
cout<<ans<<endl;
}
int main(){
IOS;
//freopen("try.in", "r", stdin);
//freopen("try2.out", "w", stdout);
//int t;cin>>t;
//while(t--)
solve();
return 0;
}
#include "iostream"
#include "cstring"
#include "string"
#include "vector"
#include "cmath"
#include "algorithm"
#include "map"
#include "set"
#include "queue"
#include "stack"
#define fi first
#define se second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define seteps(N) setprecision(N)
#define uni(x) sort(all(x)), x.erase(unique(all(x)), x.end())
#define lson (ind<<1)
#define rson (ind<<1|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 lll;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int N=1010;
const int M=1<<19;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll oone=1;
const double eps=1e-6;
const double pi=acos(-1);
int f[N][N];
void solve(){
int n,m;
cin>>n>>m;
vector<int> a(n+1,0),b(m+1,0);
rep(i,1,n+1) cin>>a[i];
rep(i,1,m+1) cin>>b[i];
mst(f,INF);
f[0][0]=0;
rep(i,1,n+1) f[i][0]=i;
rep(i,1,m+1) f[0][i]=i;
rep(i,1,n+1){
rep(j,1,m+1){
if(a[i]==b[j]) f[i][j]=min(f[i][j],f[i-1][j-1]);
f[i][j]=min(f[i][j],f[i-1][j]+1);
f[i][j]=min(f[i][j],f[i][j-1]+1);
f[i][j]=min(f[i][j],f[i-1][j-1]+1);
}
}
cout<<f[n][m]<<endl;
}
int main(){
IOS;
//freopen("try.in", "r", stdin);
//freopen("try2.out", "w", stdout);
//int t;cin>>t;
//while(t--)
solve();
return 0;
}
模板题,线段树或者树状数组都可以做
#include "iostream"
#include "cstring"
#include "string"
#include "vector"
#include "cmath"
#include "algorithm"
#include "map"
#include "set"
#include "queue"
#include "stack"
#define fi first
#define se second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define seteps(N) setprecision(N)
#define lson (ind<<1)
#define rson (ind<<1|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 lll;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int N=3*1e5+10;
const int M=1<<19;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll oone=1;
const double eps=1e-6;
const double pi=acos(-1);
int a[N];
struct Node{
int l,r,cur;
}node[N * 4];
struct SegTree{
void push_up(int ind){
node[ind].cur=node[lson].cur^node[rson].cur;
}
void push_down(int ind){
}
void build(int l,int r,int ind){
node[ind].l=l;
node[ind].r=r;
if(l==r){
node[ind].cur=a[l];
}else{
int mid=l+r>>1;
build(l,mid,lson);
build(mid+1,r,rson);
push_up(ind);
}
}
void update(int l,int r,int ind,int val){
if(l>node[ind].r || r<node[ind].l) return;
if(l<=node[ind].l && r>=node[ind].r){
node[ind].cur^=val;
}else{
push_down(ind);
update(l,r,lson,val);
update(l,r,rson,val);
push_up(ind);
}
}
void query(int l,int r,int ind,ll &ans){
if(l>node[ind].r || r<node[ind].l) return;
if(l<=node[ind].l && node[ind].r<=r){
ans^=node[ind].cur;
}else{
push_down(ind);
query(l,r,lson,ans);
query(l,r,rson,ans);
push_up(ind);
}
}
};
void solve(){
int n,q;
cin>>n>>q;
rep(i,1,n+1) cin>>a[i];
SegTree seg;
seg.build(1,n,1);
while(q--){
int op,x,y;
cin>>op>>x>>y;
if(op==1){
seg.update(x,x,1,y);
}else{
ll ans=0;
seg.query(x,y,1,ans);
cout<<ans<<endl;
}
}
}
int main(){
IOS;
//freopen("try.in", "r", stdin);
//freopen("try2.out", "w", stdout);
//int t;cin>>t;
//while(t--)
solve();
return 0;
}