Codeforces Round #298 (Div. 2)

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A - Exam 構造

 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 }

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B - Covered Path 想法題

 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 }

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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 }

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Codeforces Round #298 (Div. 2)