正睿20秋季提高十连测day2

估分： 40 + 100 + 0 = 140

实际： 10 + 100 + 0 = 110

T1：

　　错题原因：没想到正解，暴力把代码写炸了

　　正解：贪心，一层循环枚举月份，二层循环枚举买哪种钻石

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 using namespace std;
 6 
 7 typedef long long LL;
 8 
 9 const int N = 4e5 + 10, INF = 0x3f3f3f3f;
10 
11 int n, m;
12 int c[N], res;
13 
14 struct node 
15 {
16     int p, lim;
17 }a[N];
18 
19 inline bool cmp(node x, node y)
20 {
21     return x.p < y.p;
22 }
23 
24 int main()
25 {
26     scanf("%d%d", &n, &m);
27     for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
28     for (int i = 1; i <= m; i++)
29     {
30         scanf("%d%d", &a[i].p, &a[i].lim);
31         if (a[i].lim < 0)
32             a[i].lim = INF;
33     }
34 
35     sort(a + 1, a + m + 1, cmp);
36 
37     LL ans = 0;
38     for (int i = 1; i <= m; i++)
39     {
40         res = 0;
41         LL sum = 0;
42         for (int j = 1; j <= n; j++)
43         {
44             sum += a[i].lim;
45             if (sum < c[j]) res = 1, c[j] -= sum, sum = 0;
46             else sum -= c[j], c[j] = 0;
47         }
48         ans += (n * 1ll * a[i].lim - sum) * a[i].p;
49         if (!res) break;
50     }
51     printf("%lld\n", ans);
52 
53 }

View Code

T2：

　　简单的贪心排序；

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 int read()
 8 {
 9     int x = 0, f = 0;
10     char ch = getchar();
11     while (!isdigit(ch)) f = ch == '-', ch = getchar();
12     while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
13     return f ? -x : x;
14 }
15 
16 const int N = 300010;
17 
18 int n;
19 string str[N];
20 
21 bool cmp(string a, string b)
22 {
23     return a + b < b + a;
24 }
25 
26 int main()
27 {
28     n = read();
29     for (int i = 1; i <= n; i++) cin >> str[i];
30     sort(str + 1, str + 1 + n, cmp);
31     for (int i = 1; i <= n; i++) cout << str[i];
32 }

View Code

T3：

　　错题原因：不会

　　正解：题意相当于n堆石子，每次可以选择一堆或两堆拿，拿走最后一个石子的人获胜，再进行下面一番神奇的运算得出结果

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 using namespace std;
 8 
 9 typedef long long LL;
10 typedef pair<int, int> PII;
11 
12 const int mod = 1e9 + 7;
13 
14 int n;
15 int b1[60];
16 LL b2[60], b3[60], ans;
17 
18 void ins(LL x, LL y, int z)
19 {
20     for (int i = 59; i + 1; i--)
21     {
22         if (x >> i & 1 || y >> i & 1)
23         {
24             if (!b2[i] && !b3[i])
25             {
26                 b2[i] = x, b3[i] = y, b1[i] = z; 
27             }
28 
29             if (z > b1[i]) swap(x, b2[i]), swap(y, b3[i]), swap(z, b1[i]);
30 
31             LL p = b2[i], q = b3[i];
32             if ((y ^ q) >> i & 1) swap(p, q);
33             LL u = (x & ~(p | q)) | (y & p) | (~(x | y) & q);
34             LL v = (y & ~(p | q)) | (~(x | y) & p) | (x & q);
35             x = u, y = v;
36         }
37     }
38 }
39 
40 int main()
41 {
42     scanf("%d", &n);
43     for (int i = 1; i <= n; i++)
44     {
45         LL x;
46         int y;
47         scanf("%lld %d", &x, &y);
48         x ^= ans;
49         ins(x, 0, y);
50         ans = 0;
51         for (int i = 0; i < 60; i++) ans += b1[i];
52         printf("%lld\n", ans);
53     }
54 }

View Code

总结：

　　代码能力还要加强，暴力不要写炸，以及想题时多换几种角度想，第一题就一直在想第一层枚举n而导致没想到正解。第三题的类似于NIM游戏的题还要多练习，许多扩展还不会以及想不到。

posted on 2020-09-07 20:03 ArrogHie 阅读(215) 评论(0) 收藏举报