AtCoder Beginner Contest 187 ABCDE 题解

A - Large Digits

思路：签到题，读入字符串即可。

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 endl '
'
#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()
{
    string a, b;
    cin>>a>>b;
    ll sum1 = 0, sum2 = 0;
    for(int i=0; i<a.size(); i++) sum1 += a[i]-'0';
    for(int i=0; i<b.size(); i++) sum2 += b[i]-'0';
    cout<<max(sum1,sum2)<<endl;
    return 0;
}

B - Gentle Pairs

思路：暴力统计，算斜率的时候直接看分子绝对值是否小于等于分母绝对值。

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 endl '
'
#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} };

typedef struct Pos{
    ll x;
    ll y;
}P;

P a[maxn];

int main()
{
    ll n = read();
    rep(i,1,n) a[i].x = read(), a[i].y = read();
    ll sum = 0;
    rep(i,1,n) rep(j,i+1,n)
    {
        ll up = a[i].x - a[j].x;
        ll down = a[i].y - a[j].y;
        swap(up,down);
        if(abs(up)<=abs(down) ) sum++;
    }
    cout<<sum<<endl;
    return 0;
}

C - 1-SAT

思路：先Map记录，然后把有!的去掉头，看看剩下的是否存在即可。

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 endl '
'
#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();
    map<string,int> Map;
    vector<string> item;
    rep(i,1,n)
    {
        string s;
        cin>>s;
        Map[s] = 1;
        item.pb(s);
    }
    int flag = 0;
    rep(i,1,n)
    {
        if(item[i-1][0]=='!')
        {
            string t(item[i-1],1,item[i-1].size());
            if(Map[t])
            {
                cout<<t<<endl;
                flag = 1;
                break;
            }
        }
    }
    if(!flag) cout<<"satisfiable"<<endl;
    return 0;
}

D - Choose Me

思路：这题有个小wa点就是不能直接贪心拿和大的（有可能和比较小但是其中的第一个值很大）。
考虑因为要第二个人得分比第一个人高分，我们如果对一个人分析的话，第二个人最优的情况是全部都选上，就是所有的和。然后我们看要剔除尽可能的地点。
每次剔除，自身-sum，对方+第一个值， 相当于我 (-sum-第一个值)。 所以我们就按照2*第一个值 + 第二个值从小到大来排个序，然后按照这个顺序来剔除直到剩余值小于等于0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 endl '
'
#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 = 2e5+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} };

typedef struct Voter
{
    ll aoki;
    ll taka;
    ll sum;
    bool operator < (const Voter &a) const
    {
        return sum<a.sum;
    }
}V;

V a[maxn];
ll sum[maxn];

int main()
{
    ll n = read();
    rep(i,1,n) a[i].aoki = read(), a[i].taka = read(), a[i].sum = 2*a[i].aoki + a[i].taka;
    sort(a+1,a+1+n);
    ll cur = 0;
    rep(i,1,n) cur += a[i].aoki + a[i].taka;

    rep(i,1,n)
    {
        cur -= a[i].sum;
        if(cur<=0)
        {
            cout<<n-i+1<<endl;
            break;
        }
    }
    return 0;
}

E - Through Path

这题暴力不可取。因为每次询问的ab是直接相连的，所以可以从这里入手（一开始没看到是直接相连卡住了）。
我们假设1为根。若a的深度比b小的时候，也就是a是b的父节点，那么这个时候，从跟出发的所有子树中，剔除b及其子树都可以+x。
(那我们就add[1] += x， add[b] -= x， 表示1结点下面的所有子树都+x，b及其子树都-x。)
因为dfs的时候可以将父节点的增值信息传递下来，所以我们询问的时候只需要记录第一次发生改变的结点就行了。
若a是b的子树，那么这种情况只需要a的子树都+x，(即add[a] += x。)

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 endl '
'
#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 = 2e5+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} };

vector<vector<int> > D(maxn);
ll n;
ll dep[maxn];
ll add[maxn];
ll a[maxn];
ll b[maxn];

void getDep(int cur, int step, int pre)
{
    dep[cur] = step;
    for(int i=0; i<D[cur].size(); i++)
    {
        int v = D[cur][i];
        if(v!=pre) getDep(v,step+1,cur);
    }
}

void dfs(int cur, ll preSum, int pre)
{
    add[cur] += preSum;
    for(int i=0; i<D[cur].size(); i++)
    {
        int v = D[cur][i];
        if(v!=pre) dfs(v,add[cur], cur);
    }
}

int main()
{
    ll n = read();
    rep(i,1,n-1)
    {
        ll x = read();
        ll y = read();
        D[x].pb(y);
        D[y].pb(x);
        a[i] = x;
        b[i] = y;
    }
    getDep(1,1,-1);
    ll q = read();
    rep(i,1,q)
    {
        ll flag = read(), e = read(), x = read();
        if(flag==1)
        {
            if(dep[a[e]] < dep[b[e]])
            {
                add[1] += x;
                add[b[e]] -= x;
            }
            else
            {
                add[a[e]] += x;
            }
        }
        else
        {
            if(dep[b[e]] < dep[a[e]])
            {
                add[1] += x;
                add[a[e]] -= x;
            }
            else
            {
                add[b[e]] += x;
            }
        }
    }dfs(1,0,-1);
        rep(i,1,n) cout<<add[i]<<endl;
    return 0;
}