Educational Codeforces Round 11

DUANG: http://codeforces.com/contest/660

求让给定数组相邻元素互质所需要插入的数的个数并输出互质数组

不互质的中间加个1

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1111;
ll num[maxn];
ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a%b);
}
int main(){
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; ++i) scanf("%d", &num[i]);
    vector<int> k;
    for(int i = 1; i < n; ++i){
        if(gcd(num[i-1], num[i]) != 1){
            k.push_back(i);
        }
    }
    printf("%d
", k.size());
    int j = 0;
    for(int i = 0; i < n; ++i){
        if(j < k.size() && i == k[j]){
            printf("1 ");
            ++j;
        }
        printf("%d ", num[i]);
    }
    return 0;
}

给定规则

输入n行m人，按照规则坐好，再按照规则离开，输出离开顺序

瞎搞

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

const int maxn = 444;
bool vis[maxn];
int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    int a = 2*n+1, b = 1, c = 2*n+2, d = 2;
    //printf("%d %d %d %d
", a, b, c, d);
    memset(vis, true, sizeof(vis));
    for(int i = 0; i < 100; ++i){
        if(a <= m && vis[a]){
            printf("%d ", a);
            vis[a] = false;
            a += 2;
        }
        if(b <= m && vis[b]){
            printf("%d ", b);
            vis[b] = false;
            b += 2;
        }
        if(c <= m && vis[c]){
            printf("%d ", c);
            vis[c] = false;
            c += 2;
        }
        if(d <= m && vis[d]){
            printf("%d ", d);
            vis[d] = false;
            d += 2;
        }
    }
    return 0;
}

给你n个数和k个把0变成1的机会，求最多几个1并输出那种情况

！！！maple师兄指点下撸出第一发尺取

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

const int maxn = 333333;
int num[maxn];

int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < n; ++i) scanf("%d", &num[i]);
    int l = 0, r = 0, ll = 0, rr = 0, ans = 0, tmp = 0, cnt = m;
    while(r < n){
        if(num[r]){
            tmp++;
            r++;
        }
        else{
            if(cnt){
                r++;
                cnt--;
                tmp++;
            }
            else{
                if(!num[l])
                    cnt++;
                l++;
                tmp--;
            }
        }
        if(tmp > ans){
            ans = tmp;
            ll = l;
            rr = r;
        }
    }
    printf("%d
", ans);
    for(int i = 0; i < ll; ++i) printf("%d ", num[i]);
    for(int i = ll; i < rr; ++i) printf("%d ", 1);
    for(int i = rr; i < n; ++i) printf("%d ", num[i]);
    return 0;
}

给n个点求能组多少个平行四边形

题解给的想法更新了一下我的大脑 yeah~

两组能组成平行四边形的点的x坐标和相等、y坐标和相等

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

const int maxn = 2222;
int x[maxn], y[maxn];

int main(){
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; ++i) scanf("%d %d", &x[i], &y[i]);
    map<pair<int, int>, int> m;
    m.clear();
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < i; ++j)
            m[{x[i]+x[j], y[i]+y[j]}]++;
    int ans = 0;
    for(auto& p : m)
        ans += p.second*(p.second-1)/2;
    printf("%d
", ans);
    return 0;
}

for(auto& p : m) 容器遍历新姿势？

另一种想法：能构成平行四边形的一定是两个相同的变量

http://www.cnblogs.com/qscqesze/p/5373418.html

E、F..........233333333