Codeforces Round #298 (Div. 2)

A - Exam 构造

 1 #include <cstdio>
 2
 3 int ans[5000 + 5];
 4
 5 int main() {
 6     int n, cnt = 1;
 7     scanf("%d", &n);
 8     for (int i = 1; i <= n; i++) {
 9         if (i&1) ans[i] = cnt;
10         else {
11             ans[i] = n+1-cnt;
12             cnt++;
13         }
14     }
15     ans[0] = ans[n];
16     if (n == 1 || n == 2) {
17         printf("1\n1\n");
18         return 0;
19     }
20     if (n == 3) {
21         printf("2\n1 3\n");
22         return 0;
23     }
24     printf("%d\n", n);
25     for (int i = 0; i < n; i++)
26         printf("%d ", ans[i]);
27     return 0;
28 }

B - Covered Path 想法题

 1 #include <cstdio>
 2
 3 inline int min(int a, int b) {
 4     return a < b ? a : b;
 5 }
 6
 7 int main() {
 8     int v1, v2, t, d;
 9     scanf("%d%d%d%d", &v1, &v2, &t, &d);
10     int ans =  0;
11     for (int i = 0; i < t; i++)
12         ans += min(v1+i*d, v2+(t-i-1)*d);
13     printf("%d\n", ans);
14     return 0;
15 }

C - Polycarpus‘ Dice 求上下界

 1 #include <cstdio>
 2
 3 const int N = 100000*2 + 5;
 4 typedef long long ll;
 5 ll ans[N], num[N], suf[N], pre[N];
 6
 7 int main() {
 8     int n;
 9     ll s;
10     scanf("%d%lld", &n, &s);
11     for (int i = 1; i <= n; i++)
12         scanf("%lld", num+i);
13     for (int i = 1; i <= n; i++)
14         pre[i] = pre[i-1] + num[i];
15     for (int i = n; i >= 1; i--)
16         suf[i] = suf[i+1] + num[i];
17     ll can_min, can_max;
18     for (int i = 1; i <= n; i++) {
19         can_min = s - pre[i-1] - suf[i+1];
20         if (can_min < 1) can_min = 1;
21         can_max = s - n + 1;
22         if (can_max > num[i]) can_max = num[i];
23         ans[i] = num[i] - (can_max - can_min + 1);
24     }
25     for (int i = 1; i <= n; i++)
26         printf("%lld ", ans[i]);
27     return 0;
28 }