Codeforces Round #345 (Div. 2)题解

A. Joysticks

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模拟就行了，
每次都先给电量少的充电，直到另一个的电量不足以多使用1minute，

#include <bits/stdc++.h>#define ll long long#define N 100using namespace std;int main(){#ifndef ONLINE_JUDGE//  freopen("1.txt", "r", stdin);#endif    int i, j, k;    int a1, a2, ans = 0, t;    cin >> a1 >> a2;    while(a1 > 0 && a2 > 0)    {        if (a1 < a2)    swap(a1, a2);        t = a1/2;        if (a1 == 2)        {            ans++;            break;        }        if (a1&1)        {            a2 += t;            a1 = 1;        }        else        {            t--;            a1 = 2;            a2 += t;        }        ans += t;        if (a1 < 2 && a2 < 2)   break;    }    cout << ans;    return 0;}

B. Beautiful Paintings

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set详解

#include <bits/stdc++.h>#define ll long long#define N 1010using namespace std;set<int> s[N];int main(){#ifndef ONLINE_JUDGE//  freopen("1.txt", "r", stdin);#endif    int i, j, k, t;    int n, ans;    for (i = 0; i < 1000; i++)  s[i].clear();    cin >> n;    for (i = 0; i < n; i++)    {        cin >> t;        for (j = 0; j < 1000; j++)        {            if (s[j].find(t) == s[j].end())            {                s[j].insert(t);                break;            }        }    }    ans = 0;    for (i = 0; i < 1000; i++)    {        t = s[i].size();        ans += t?t-1:0;    }    cout << ans;    return 0;}

C. Watchmen

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map详情

这道题竟然在终测的时候爆int了 /(ㄒoㄒ)/~~
map<int, int>应该改成map<long long, long long>，这样直接就AC了
这道题的意思就是找有多少对坐标对满足曼和顿距离和欧几里得距离相等，满足这种条件的坐标对都在同一条直线上。

用map计算就行了

#include <bits/stdc++.h>#define ll long long#define N   200010using namespace std;struct point{    int x, y;}node[N];bool cmp_x(point a, point b){    if (a.x == b.x) return a.y < b.y;    return a.x < b.x;}map<ll,ll> mx, my;int main(){#ifndef ONLINE_JUDGE    freopen("1.txt", "r", stdin);#endif    int i, j, k, n, xx, yy;    ll ans = 0, t, tmp;    map<ll,ll>::iterator it;    scanf("%d", &n);    for (i = 0; i < n; i++)    {        scanf("%d%d", &node[i].x, &node[i].y);        if (mx.find(node[i].x) == mx.end())        {            mx.insert(pair<ll,ll>(node[i].x, 1));        }        else        {            mx[node[i].x]++;        }        if (my.find(node[i].y) == my.end())        {            my.insert(pair<ll,ll>(node[i].y, 1));        }        else        {            my[node[i].y]++;        }    }    for (it = mx.begin(); it != mx.end(); it++)        ans += (it->second)*((it->second)-1)/2;    for (it = my.begin(); it != my.end(); it++)        ans += it->second*(it->second-1)/2;    sort(node, node+n, cmp_x);    t = 1;    for (i = 1; i < n; i++)    {        if (node[i].x == node[i-1].x && node[i].y == node[i-1].y)        {            t++;        }        else        {            ans -= t*(t-1)/2;            t = 1;        }    }    ans -= t*(t-1)/2;    cout << ans;    return 0;}

D. Image Preview

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#include <bits/stdc++.h>#define ll long long#define N   500010using namespace std;int n, a, b, t;int num[N]; //前i个数有多少需要旋转的 int is[N];int main(){#ifndef ONLINE_JUDGE//  freopen("1.txt", "r", stdin);#endif    char c;    scanf("%d%d%d%d", &n, &a, &b, &t);    getchar();    for (int i = 0; i < n; i++)    {        scanf("%c", &c);        is[i] = (c == 'w');        num[i] = is[i];        if (i)  num[i] += num[i-1];    }    if (is[0]*b+1 > t)      {        cout << 0;        return 0;    }    int l = 0, sum = 0; //从第一个想第n个开始看     for (; ; )    {        if (is[l])  sum += b;        sum++;        if (sum > t)    break;        sum += a;        l = (l-1+n)%n;        if (!l) break;  //全部都已经看完     }//如果一直向右看，可以看到l张，共看n-1-l+1张照片     if (!l)    {        cout << n;        return 0;    }    int ws, ans = n-1-l+1;//ws为需要旋转的照片的数量    for (int r = 1; r < n; r++)    {        ws = num[r];//0到r里面需要翻转的数量        ws += num[n-1] - num[l];//l到n-1里面需要翻转的数量        while(l && ws*b+r+1+(n-1-l)+a*(r+(n-1-l))+a*min(r, n-1-l) > t)        {            l = (l+1)%n;            ws = num[r];            ws += num[n-1]-num[l];        }        if (!l) break;        ans = max(ans, r+1+n-1-l);    }    cout << ans;    return 0;}