Codeforces Round #478 (Div. 2)

掉分场，第四题应该细心下来算一下样例

A. Aramic script

题目大意：给定n个字符串，统计有多少不同的字符串，字符串不同当且仅当出现的字符（不考虑次数）不同，只有小写字母。
题解：hash一下，丢进Set里

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define CLR(a) memset(a, 0, sizeof(a))
#define DBG(x) cout<<(#x)<<"="<<x<<endl
#define FOR(i, a, b)  for(int i=(a); i<(b); i++)
#define REP(i, a, b)  for(int i=(a); i<=(b); i++)
#define DOWN(i, a, b) for(int i=(a); i>=(b); i--)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1000000009;
const int N= 5e4 +10;

int n;
ll t;
int len;
set<ll> S;
string s;
ll w[30], vis[30];

int main() {
    cin>>n;
    w[0]=1;
    FOR(i, 1, 26) w[i]=w[i-1]*2;
    REP(i, 1, n) {
        cin>>s;
        len=s.length();
        memset(vis, 0, sizeof(vis));
        FOR(j, 0, len) {
            vis[s[j]-'a']=1;
        }
        t=0;
        FOR(j, 0, 26) {
            t+=vis[j]*w[j];
        }
        S.insert(t);
    }
    cout<<S.size()<<endl;
    //cout<<1.*clock()/CLOCKS_PER_SEC<<"ms"<<"\n";
    return 0;
}

B. Mancala

题目大意：14个洞，每个洞有一些奇数或偶数个石头，一次操作为拿出一个洞内所有石头，逆时针一个洞一个石头分配，直到分完，最后得分为，偶数个石头的洞上的石头个数
题解：模拟一下，需要注意循环的处理，一般是将序列延长一倍，还需要注意的是分配石头的时候是向上整除

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define CLR(a) memset(a, 0, sizeof(a))
#define DBG(x) cout<<(#x)<<"="<<x<<endl
#define FOR(i, a, b)  for(int i=(a); i<(b); i++)
#define REP(i, a, b)  for(int i=(a); i<=(b); i++)
#define DOWN(i, a, b) for(int i=(a); i>=(b); i--)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1000000009;
const int N= 5e4 +10;

ll a[30], b[30];
ll score;
ll ans=0;

int main() {
    REP(i, 1, 14) {
        cin>>a[i];
    }
    REP(i, 15, 27) {
        a[i]=a[i-14];
    }
    REP(i, 1, 14) {
        if (!a[i]) continue;
        memset(b, 0, sizeof(b));
        REP(j, 1, 14) {
            b[i+j]=(a[i]-(j-1)+13) / 14;
        }
        score=0;
        REP(j, 1, 13) {
            if ((b[i+j]+a[i+j])%2==0) score+=b[i+j]+a[i+j];
        }
        if (b[i+14]%2==0) score+=b[i+14];
        ans=max(score, ans);
    }
    cout<<ans<<endl;

    //cout<<1.*clock()/CLOCKS_PER_SEC<<"ms"<<"\n";
    return 0;
}

C. Valhalla Siege

题目大意：n个勇士，每个勇士有一个防御值，q轮攻击（从第一个勇士开始攻击），每轮攻击有一个攻击值（可累计），问每轮攻击后剩下多少人，当全部勇士在某轮阵亡时，全部勇士将在这轮结束时全体复活，询问每轮结束还剩多少勇士.
题解：预处理n个勇士的前缀和，每轮攻击累加，二分出还剩多少勇士，若不剩勇士，则将总攻击值置0，输出n

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define CLR(a) memset(a, 0, sizeof(a))
#define DBG(x) cout<<(#x)<<"="<<x<<endl
#define FOR(i, a, b)  for(int i=(a); i<(b); i++)
#define REP(i, a, b)  for(int i=(a); i<=(b); i++)
#define DOWN(i, a, b) for(int i=(a); i>=(b); i--)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1000000009;
const int N= 2e5 +10;

int n, q;
ll sum[N], a[N], k[N];
ll arr;
ll pos;

int main() {
    scanf("%d %d", &n, &q);
    REP(i, 1, n) {
        scanf("%lld", &a[i]);
        sum[i]=sum[i-1]+a[i];
    }
    REP(i, 1, q) {
        scanf("%lld", &k[i]);
    }

    sum[n+1]=LL_INF;
    REP(i, 1, q) {
        arr+=k[i];
        pos=lower_bound(sum+1, sum+n+1, arr) - sum;
        if (pos==n+1) {
            cout<<n<<endl;
            arr=0;
            continue;
        }
        if ( (pos==n&&sum[pos]==arr) ) {
            cout<<n<<endl;
            arr=0;
            continue;
        }
        if (sum[pos]==arr) printf("%lld\n", n-pos);
        else printf("%lld\n", n-pos+1);
    }


    //cout<<1.*clock()/CLOCKS_PER_SEC<<"ms"<<"\n";
    return 0;
}

D. Ghosts

题目大意：有n个幽灵，每个幽灵有一个起点，起点都在一条线上（很关键），给出分速度vx，vy，问时间在($-\infty$, $+\infty$) 他们相互之间两两相遇的次数。
题解：背景是几何，实则发现约束条件很多，经过物理公式推导，只要满足vy-a$\cdot$vx相等且vx不相等，幽灵便会相遇

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define CLR(a) memset(a, 0, sizeof(a))
#define DBG(x) cout<<(#x)<<"="<<x<<endl
#define FOR(i, a, b)  for(int i=(a); i<(b); i++)
#define REP(i, a, b)  for(int i=(a); i<=(b); i++)
#define DOWN(i, a, b) for(int i=(a); i>=(b); i--)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1000000009;
const int N= 2e5 +10;

int n, a, b;
ll x, vx, vy;
ll ans;
map<ll, ll> m;
map<pll, ll> m2;

int main() {
    cin>>n>>a>>b;
    REP(i, 1, n) {
        scanf("%lld %lld %lld", &x, &vx, &vy);
        ans+=(m[vy-a*vx]-m2[{vy-a*vx,vx}])*2;
        m2[{vy-a*vx,vx}]++;
        m[vy-a*vx]++;
    }
    cout<<ans<<endl;
    //cout<<1.*clock()/CLOCKS_PER_SEC<<"ms"<<"\n";
    return 0;
}

教练，我想打ACM