A. Mezo Playing Zoma
题意：给你一个向右向左的指令，每个指令可以朝那个方向走一个单位，问你可以随意选出子序列来走，那可能到达的点有多少个。
思路：从范围上考虑就秒了。看最左和最右能到哪，然后加上原点就是所有可能的点数
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
int main()
{
ll n = read();
string s;
cin>>s;
int l = 0, r = 0;
for(int i=0; i<s.size(); i++)
{
if(s[i] == 'L') l++;
else r++;
}
cout<<l+r+1<<endl;
return 0;
}
B. Just Eat It!
题意：给你一个序列，问你所有的和是否严格大于所有子段（不包含[1,n]这个）的和。
思路：其实就是模板题。求一个最大字段和，同时记录开始和结束位置。若存在一个满足限制条件的最大字段和dp[i]，则输出NO，反之输出YES。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll dp[maxn];
ll a[maxn];
ll s[maxn];
ll t[maxn];
int main()
{
int kase;
cin>>kase;
while(kase--)
{
ll n = read();
rep(i,0,n+1) dp[i] = 0;
ll sum = 0;
rep(i,1,n) a[i] = read(), sum += a[i];
ll st = 1, ed = 1;
rep(i,1,n)
{
if(dp[i-1] + a[i] > a[i])
{
dp[i] = dp[i-1] + a[i];
ed = i;
s[i] = st, t[i] = ed;
}
else
{
dp[i] = a[i];
s[i] = t[i] = st = ed = i;
}
}
int flag = 0;
rep(i,1,n) if(dp[i]>=sum&&!(s[i]==1&&t[i]==n)) {flag=1; break;}
puts(flag?"NO":"YES");
}
return 0;
}
C. Fadi and LCM
题意：问你找到一对(a,b),在满足 LCM(a,b) = X 的情况下找出max(a,b)最小的一对。
由于(LCM(a,b)) = (abover gcd(a,b))，所以当(gcd(a,b))=1的时候能把ab限制的最小。所以只需要枚举一遍[1,(sqrt{x})]找到max(a,b)最小的对即可。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
int main()
{
ll x = read();
ll m = sqrt(x*1.0);
ll ansx = x, ansy = 1;
rep(i,1,m)
{
if(x%i) continue;
ll j = x/i;
if(gcd(i,j)==1&&max(i,j)<max(ansx,ansy)) ansx = i, ansy = j;
}
cout<<ansx<<' '<<ansy<<'
';
return 0;
}