AtCoder Beginner Contest 129

文章目录

AtCoder Beginner Contest 129
- A
- B
- C
- D

A

水题

int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int a, b, c;cin >> a>> b >> c;int ans = a+ b;ans = min(ans ,a + c);ans = min(ans, b + c);cout << ans << endl;return 0;
}

B

前缀和

const int MAXN = 105;
int a[MAXN];
int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int  n;cin >> n;int ans = 1e9+5;for(int i = 1; i <= n; i++) cin >> a[i];for(int i = 1; i <= n; i++) a[i] += a[i-1];for(int i = 1; i <= n; i++){ans = min(ans, int(abs(a[i] - a[n] + a[i])));}cout << ans << endl;return 0;
}

C

变种fibonacci

const int MAXN = 1e5+5;
int vis[MAXN];
ll f[MAXN];
const int mod = 1e9+7;
int main()
{//FIN;ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int n;cin >> n;int m;cin >> m;for(int i = 0; i < m; i++){int num;cin >> num;vis[num] = 1;}if(!vis[1]) f[1] = 1;f[0] = 1;for(int i = 2; i <= n; i++){if(!vis[i-1]) f[i] += f[i-1];if(!vis[i-2]) f[i] += f[i-2];f[i] %= mod;}cout << f[n] << endl;return 0;
}

D

比赛的时候抽风
先预处理即可

string s[2010];
int a[2010][2010], b[2010][2010];
int numa[2010 * 2010], numb[2010 * 2010];
int main()
{int i, j, k, n, m, x, y, h = 0;scanf("%d%d", &n, &m);for (i = 0; i < n; i++){cin >> s[i];}for (i = 0; i < n; i++){x = 0;for (j = 0; j < m; j++){if (s[i][j] == '#'){if (x > 0){numa[h] = x;x = 0;h++;}}else{a[i][j] = h;x++;}}if (x > 0){numa[h] = x;h++;}}h = 0;for (i = 0; i < m; i++){x = 0;for (j = 0; j < n; j++){if (s[j][i] == '#'){if (x > 0){numb[h] = x;x = 0;h++;}}else{b[j][i] = h;x++;}}if (x > 0){numb[h] = x;h++;}}int ma = 0;for (i = 0; i < n; i++){for (j = 0; j < m; j++){ma = max(ma, numa[a[i][j]] + numb[b[i][j]] - 1);}}printf("%d", ma);return 0;
}

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11.25 AtCoder Beginner Contest 129

AtCoder Beginner Contest 129

文章目录

A

B

C

D

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