A - Weird Function
B - Longest Segment
C - Happy New Year!
D - Prefix K-th Max
E - Arithmetic Number
F - Reordering
G - Divide a Sequence
写个函数
int f(int x){return x*x+2*x+3;}
int main()
{ int t;scanf("%d", &t);cout<<f(f(f(t)+t)+f(f(t)));return 0;
}
暴力计算
int x[N], y[N];
int main()
{ int n, ma = 0;scanf("%d", &n);for(int i = 1;i <= n;i ++){scanf("%d%d", x+i, y+i);for(int j = 1;j < i;j ++)ma = max(ma, (x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));}printf("%.6lf", (double)sqrt(ma*1.0));return 0;
}
可以看作是二进制数，0 - > 0, 1 - > 2
string ans;
int main()
{ LL k;scanf("%lld", &k);while(k) ans += (k&1 ? '2' : '0'), k >>= 1;reverse(ans.begin(), ans.end());cout<<ans<<endl;return 0;
}
树状数组+二分O(nlog2n)O(nlog^2n)O(nlog2n)
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
int tr[N], n, k;
void add(int u){while(u <= n)tr[u]++, u += lowbit(u);return;}
int quary(int u){int ans = 0;while(u)ans += tr[u], u -= lowbit(u);return ans;}int main()
{ scanf("%d%d", &n, &k);for(int i = 1, x;i <= n;i ++){scanf("%d", &x);add(x);if(i >= k){int t = i-k+1, l = 1, r = n;while(l < r)quary(mid) >= t ? r = mid : l = mid+1;cout<<l<<endl;}}return 0;
}
暴力dfs
LL x, ans = 1e18;LL se(int i, int d, LL sum)
{if(sum >= x) return sum;if(i+d >= 10 || i+d < 0) return 1e18;return se(i+d, d, sum*10+i+d);
}int main()
{ scanf("%lld", &x);for(int i = 1;i <= 9;i ++)for(int j = -8;j <= 9;j ++)ans = min(ans, se(i, j, i));cout<<ans<<endl;return 0;
}
因为是作为排列那么顺序就没必要了，考虑 dp dp[i][j]表示前i种字母构成长度为j的数量有多少那么dp[i][j]=dp[i−1][j]+dp[i−1][j−1]∗Cj1.....+dp[i−1][j−x]∗Cjx,x<=num[i]并且x<=jdp[i][j]表示前i种字母构成长度为j的数量有多少那么dp[i][j] = dp[i-1][j] + dp[i-1][j-1]*C_{j}^{1} ..... +dp[i-1][j-x]*C_{j}^{x}, x <= num[i] 并且 x <= jdp[i][j]表示前i种字母构成长度为j的数量有多少那么dp[i][j]=dp[i−1][j]+dp[i−1][j−1]∗Cj1.....+dp[i−1][j−x]∗Cjx,x<=num[i]并且x<=j 总复杂度O(26n2)O(26n^2)O(26n2)
void mull(int &a, LL b){a = a*b%mod;return ;}
void add(int &a, LL b){a = (a+b)%mod;return ;}int ans[5100];
int num[26];
string s;LL inv[N],in[N],finv[N];
void ini(int n,LL p)
{inv[0]=inv[1]=in[1]=in[0]=finv[0]=finv[1]=1;for(int i=2;i<n;i++)finv[i]=(p-p/i)*finv[p%i]%p,in[i]=in[i-1]*i%p,inv[i]=inv[i-1]*finv[i]%p;
}
int C(int a,int b){return in[a]*inv[a-b]%mod*inv[b]%mod;}
int A(int a,int b){return in[a]*inv[a-b]%mod;}int main()
{ ini(5100, mod);cin>>s;for(auto x:s) num[x-'a']++;ans[0] = 1;for(int i = 0;i < 26;i ++)for(int j = 5000;j >= 0;j --)for(int k = 1;k <= num[i];k ++)if(j >= k)add(ans[j], (LL)ans[j-k]*C(j, k));ans[0] = 0;for(int i = 1;i <= 5000;i ++)add(ans[0], ans[i]);cout<<ans[0]<<endl;return 0;
}
考虑dp, dp[i] 表示前i个数字构成的答案，dp[i]=dp[i−1]∗(max(i−>i)−min(i−>i)+dp[i−2]∗(max(i−1−>i)−min(i−1−>i))+......dp[i] = dp[i-1]*(max(i->i)-min(i->i) + dp[i-2]*(max(i-1 -> i) - min(i-1 -> i)) + ......dp[i]=dp[i−1]∗(max(i−>i)−min(i−>i)+dp[i−2]∗(max(i−1−>i)−min(i−1−>i))+...... 从这里可以再化简，把 dp[i−x]∗(max−min)=dp[i−x]∗max−dp[i−x]∗min然后可以看出max(i,i)<=max(i−1,i)<=max(i−2,i)<=.....dp[i-x]*(max-min) = dp[i-x]*max - dp[i-x]* min然后可以看出max(i,i) <= max(i-1,i)<=max(i-2,i)<=.....dp[i−x]∗(max−min)=dp[i−x]∗max−dp[i−x]∗min然后可以看出max(i,i)<=max(i−1,i)<=max(i−2,i)<=.....那么这个把末尾新加一个数的操作可以用单调栈来维护，一个单调递增栈，一个单调递减栈。 注：代码很难看
void mull(int &a, LL b){a = a*b%mod;return ;}
void add(int &a, LL b){a = (a+b)%mod;return ;}stack<PII>st, lt;
int main()
{ int n, ans = 0, x; // ans 代表 dp[i-1] scanf("%d%d", &n, &x);st.push({x, 1}); // 初始状态吧，不加这个好像不会定义初始的ans了。 lt.push({x, 1}); for(int i = 2;i <= n;i ++){scanf("%d", &x);int sum = ans, t = ans;while(st.size() && st.top().first <= x)add(sum, st.top().second), add(ans, -(LL)st.top().first*st.top().second%mod), st.pop();st.push({x, sum});add(ans, (LL)x*sum);sum = t;while(lt.size() && lt.top().first >= x)add(sum, lt.top().second), add(ans, (LL)lt.top().first*lt.top().second%mod), lt.pop();lt.push({x, sum});add(ans, -(LL)x*sum);}add(ans, mod);cout<<ans<<endl;return 0;
}
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