Codeforces Round #365 (Div. 2) （703A，703B（容斥），703C（几何），703D（树状数组））

Mishka and Game

题目链接：

http://codeforces.com/problemset/problem/703/A

解题思路：

In this problem you had to do use the following algo. If Mishka wins Chris in the current round, then increase variable countM by 1. Otherwise (if Chris wins Mishka) increase variable countC. After that you had to compare this values and print the answer.

AC代码：

#include <bits/stdc++.h>
using namespace std;

int main(){
    int n;
    while(~scanf("%d",&n)){
        int cnt1 = 0,cnt2 = 0;
        int x,y;
        for(int i = 0; i < n; ++i){
            scanf("%d%d",&x,&y);
            if(x > y)
                ++cnt1;
            else if(x < y)
                ++cnt2;
        }
        if(cnt1 > cnt2)
            puts("Mishka");
        else if(cnt1 < cnt2)
            puts("Chris");
        else
            puts("Friendship is magic!^^");
    }
    return 0;
}

Mishka and trip

题目链接：

http://codeforces.com/problemset/problem/703/B

解题思路：

Let's look at the first capital. Note that the total cost of the outgoing roads is c[id1] · (sum - c[id1]), where sum — summary beauty of allcities. Thus iterating through the capitals we can count the summary cost of roads between capitals and all the other cities. But don't forget that in this case we count the roads between pairs of capitals twice. To avoid this on each step we should update sum = sum - c[idcur] , where idcur is the position of current capital. In the end we should add to the answer the cost of roads between "non-capital" neighbour cities.

Complexity — O(n).

算法思想：

题解是O（nlogn）的，其实我们只需要求出前缀和，然后先求出每个首都城市x，对答案的贡献：c[x] * others即可。O（n）.

AC代码：

#include <bits/stdc++.h>
using namespace std;

typedef long long int ll;
const int N = 100005;
int c[N];
bool vis[N];

int main(){
    int n,k;
	while(~scanf("%d%d",&n,&k)){
        memset(vis,false,sizeof(vis));
        ll sum = 0;
        for(int i = 1; i <= n; ++i){
            scanf("%d",&c[i]);
            sum += c[i];
        }
        ll visc = 0,ans = 0;
        for(int i = 1; i <= k; ++i){
            int x;
            scanf("%d",&x);
            visc += c[x];
            ans += c[x]*(sum-visc);
            vis[x] = true;
        }
        for(int i = 1; i <= n; ++i){
            if(vis[i] || vis[i+1])
                continue;
            ans += c[i]*c[i+1];
        }
        if(!vis[1] && !vis[n])
            ans += c[1]*c[n];
        printf("%lld\n",ans);
	}
	return 0;
}

Chris and Road

题目链接：

http://codeforces.com/problemset/problem/703/C

解题思路：

Imagine that the bus stands still and we move "to the right" with a constant speed v. Then it's not hard to see that movement along the liney = (u / v) · (x  -  t0) is optimal, where t0 — time in which we begin our movement. In this way answer is t = t0 + (w / u).

If t0 = 0, then we start our movement immediately. In this case we need to check that our line doesn't intersect polygon (either we can cross the road in front of a bus, or the bus is gone).

Otherwise we need to find such minimal t0 that our line is tangent to the polygon. It can be done with binary search.

Complexity — O(n log n).

算法思想：

我们可以先考虑人直接过去，而车撞不到人的情况；然后再考虑人等车先过，人再过的情况即可。

AC代码：

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 10005;
int n,w,v,u;
int x[N],y[N];

bool pre(){
    for(int i = 0; i < n; ++i)
        if(ll(x[i])*u < ll(y[i])*v)
            return false;
    return true;
}

double solve(){
    double ret = w*1.0/u;
    for(int i = 0; i < n; ++i)
        ret = max(ret, x[i]*1.0/v + (w-y[i])*1.0/u);
    return ret;
}

int main(){
    while(~scanf("%d%d%d%d",&n,&w,&v,&u)){
        for(int i = 0; i < n; i++)
            scanf("%d%d",&x[i],&y[i]);
        if(pre()){
            printf("%.6lf\n",w*1.0/u);
            return 0;
        }
        printf("%.6lf\n",solve());
    }
    return 0;
}

Mishka and Interesting sum

题目链接：

http://codeforces.com/problemset/problem/703/D

解题思路：

Easy to see, that the answer for query is XOR-sum of all elements in the segment xored with XOR-sum of distinct elements in the segment. XOR-sum of all numbers we can find in O(1) using partial sums. As for the XOR-sum of distinct numbers... Let's solve easier problem.

Let the queries be like "find the number of distinct values in a segment". Let's sort all the queries according to their right bounds and iterate through all elements of our array. We also need to make a list last, where last[value] is the last position of value on the processed prefix of array. Assume we are at position r. Then the answer for the query in the segment [l,  r] (l ≤ r) is , where cnt[i] = 1 iflast[ai] = i and 0 otherwise. It's easy to store and update such values in cnt. When moving to the next position we have to make the following assignments: cnt[last[ai]] = 0, cnt[i] = 1, last[ai] = i. To get described sum in O(log n) we can use segment tree (or Fenwick tree) instead of standard array.

Now let's turn back to our problem. Everything we have to do is to change assignment cnt[i] = 1 to cnt[i] = ai and count XOR-sum instead of sum. Now we can solve this problem in O(n log n).

题目大意：

给你n个数，然后m次查询，每一次查询输出所在给定区间内出现偶数次数的数的异或值，如果只有一个出现偶数次数的数，则直接

输出该数，如果没有出现偶数次数的数，则直接输出0。

算法思想：

和求一个区间内不同元素个数的做法一样。树状数组存的是不同元素的前缀异或和。用map标记数a[i]最近的出现的下标。其实就是

先求出这个区间的（异或和），然后再异或上这个区间不同数的异或和（即由树状数组来完成的）。

AC代码：

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1000005;
map<int,int> mp;
vector<int> que[N];
int n,m;
int a[N];
int l[N],r[N];
ll res[N],bit[N];
ll ans[N];

int lowbit(int x){
    return x&(-x);
}

void update(int x, ll val){
	for(int i = x; i <= n; i += lowbit(i))
		bit[i] ^= val;
}

int sum(int x){
	ll res = 0;
	for(int i = x; i > 0; i -= lowbit(i))
		res ^= bit[i];
	return res;
}

int main(){
	while(~scanf("%d",&n)){
        mp.clear();
        memset(res,0,sizeof(res));
        for(int i = 0; i <= n; ++i)
            que[i].clear();
        for(int i = 1; i <= n; ++i){
            scanf("%d",&a[i]);
            res[i] = (res[i-1]^a[i]);
        }
        scanf("%d",&m);
        for(int i = 0; i < m; ++i){
            scanf("%d%d",&l[i],&r[i]);
            que[l[i]].push_back(i);
            ans[i] = res[r[i]]^res[l[i]-1];
        }
        for(int i = n; i >= 1; --i){
            int x = a[i];
            if(mp.count(x))
                update(mp[x],x);
            update(i,x);
            mp[x] = i;
            for(int j = 0; j < que[i].size(); ++j){
                int id = que[i][j];
                ans[id] ^= sum(r[id]);
            }
        }
        for(int i = 0; i < m; ++i)
            printf("%lld\n", ans[i]);
	}
	return 0;
}