Codeforces Round #481 (Div. 3)

题解:枚举公差,a[2]的操作有三种情况:a[2]+1,a[2],a[2]-1;a[1]的操作有三种情况:a[1]+1,a[1],a[1]-1。
 1 #pragma warning(disable:4996)
 2 #include<string>
 3 #include<queue>
 4 #include<cstdio>
 5 #include<cstring>
 6 #include<iostream>
 7 #include<algorithm>
 8 #define ll long long
 9 using namespace std;
10 
11 const int maxn = 100005;
12 const int INF = 1e9 + 7;
13 
14 int n;
15 int a[maxn];
16 
17 int cal(int d,int last) {
18     int cnt = 0;
19     for (int i = 3; i <= n; i++) {
20         if (a[i] - last == d) {
21             last = a[i];
22         }
23         else if (a[i] - 1 - last == d) {
24             cnt++;
25             last = a[i] - 1;
26         }
27         else if (a[i] + 1 - last == d) {
28             cnt++;
29             last = a[i] + 1;
30         }
31         else return -1;
32     }
33     return cnt;
34 }
35 
36 int main()
37 {
38     while (cin >> n) {
39         for (int i = 1; i <= n; i++) cin >> a[i];
40         if (n == 1 || n == 2) cout << "0" << endl;
41         else {
42             int ans = INF;
43 
44             int d1 = a[2] + 1 - a[1];
45             int last = a[2] + 1;
46             int x = cal(d1, last);
47             if (x != -1) ans = min(ans, x + 1);
48             
49             d1 = a[2] + 2 - a[1];
50             last = a[2] + 1;
51             x = cal(d1, last);
52             if (x != -1) ans = min(ans, x + 2);
53 
54             d1 = a[2] - (a[1] + 1);
55             last = a[2];
56             x = cal(d1, last);
57             if (x != -1) ans = min(ans, x + 1);
58 
59             d1 = a[2] - (a[1] - 1);
60             last = a[2];
61             x = cal(d1, last);
62             if (x != -1) ans = min(ans, x + 1);
63 
64             d1 = a[2] - 1 - (a[1] + 1);
65             last = a[2] - 1;
66             x = cal(d1, last);
67             if (x != -1) ans = min(ans, x + 2);
68 
69             d1 = a[2] - 1 - a[1];
70             last = a[2] - 1;
71             x = cal(d1, last);
72             if (x != -1) ans = min(ans, x + 1);
73 
74             d1 = a[2] - a[1];
75             last = a[2];
76             x = cal(d1, last);
77             if (x != -1) ans = min(ans, x);
78             
79             d1 = (a[2] + 1) - (a[1] + 1);
80             last = a[2] + 1;
81             x = cal(d1, last);
82             if (x != -1) ans = min(ans, x + 2);
83 
84             d1 = (a[2] - 1) - (a[1] - 1);
85             last = a[2] - 1;
86             x = cal(d1, last);
87             if (x != -1) ans = min(ans, x + 2);
88 
89             if (ans == INF) {
90                 cout << "-1" << endl;
91             }
92             else cout << ans << endl;
93         }
94     }
95     return 0;
96 }

E. Bus Video System

题解:前缀和表示当前这个车站相对于起点(起点车站标为:0)下了多上人或者上了多少人。那么Max和Min分别表示极限情况。Min说明起点至少有多少人(abs(Min)),Max说明起点至多能有多少人(m-Max)。所以答案就是(m-Max)-abs(Min)+1。

 1 #pragma warning(disable:4996)
 2 #include<string>
 3 #include<queue>
 4 #include<cstdio>
 5 #include<cstring>
 6 #include<iostream>
 7 #include<algorithm>
 8 #define ll long long
 9 using namespace std;
10 
11 const int maxn = 1005;
12 
13 int n, m;
14 
15 int main()
16 {
17     while (cin >> n >> m) {
18         int res = 0, Max = 0, Min = 0;
19         for (int i = 1; i <= n; i++) {
20             int d;
21             cin >> d;
22             res += d;
23             Max = max(Max, res);
24             Min = min(Min, res);
25         }
26         int ans = m - Max + min(0, Min) + 1;
27         if (ans < 0) {
28             cout << 0 << endl;
29         }
30         else {
31             cout << ans << endl;
32         }
33     }
34     return 0;
35 }

 F. Mentors

题解:哈希后求前缀和,可以得到小于当前数的有多少个,然后统计争吵中小于当前数的有多少个。

感受:这题没多大意思,因为给出的争吵关系不具有传递性~~~,如果有传递性就灰常麻烦了,我还多写了并查集 :( 。

#pragma warning(disable:4996)
#include<map>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 200005;

int n, m;
int pa[maxn], d[maxn], c[maxn], cnt[maxn], ans[maxn];

map<int, int> p;

int Find(int a) {
    if (a == pa[a]) return a;
    return pa[a] = Find(pa[a]);
}

void Union(int a, int b) {
    int x = Find(a);
    int y = Find(b);
    if (x != y) pa[y] = x;
}

void Inite() {
    p.clear();
    memset(cnt, 0, sizeof(cnt));
    memset(ans, 0, sizeof(ans));
    for (int i = 1; i <= n; i++) pa[i] = i;
}

int main()
{
    while (scanf("%d%d", &n, &m) != EOF) {

        Inite();
        for (int i = 1; i <= n; i++) {
            scanf("%d", &c[i]);
            d[i] = c[i];
        }

        sort(d + 1, d + n + 1);
        int k = unique(d + 1, d + n + 1) - (d + 1);
        for (int i = 1; i <= k; i++) p[d[i]] = i;

        for (int i = 1; i <= m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            if (c[u] > c[v]) ans[u]++;
            if (c[u] < c[v]) ans[v]++;
        }

        cnt[0] = 0;
        for (int i = 1; i <= n; i++) cnt[p[c[i]]]++;
        for (int i = 1; i <= k; i++) cnt[i] += cnt[i - 1];

        for (int i = 1; i <= n; i++) {
            cout << cnt[p[c[i]] - 1] - ans[i] << endl;
        }

    }
    return 0;
}

 

 

 

 

 

 

 

 

 

 

 

 

posted @ 2018-05-13 21:53  天之道,利而不害  阅读(269)  评论(0编辑  收藏  举报