做网站的问卷调查/优化标题关键词技巧

 快速排序

快速排序

#include <iostream>
#include <algorithm>
using namespace std;
int n;
int s[100000];
int main()
{cin>>n;for(int i=0;i<n;i++){cin>>s[i];}sort(s,s+n);for(int i=0;i<n;i++){cout<<s[i]<<" ";}cout<<endl;return 0;
}

第K个数

#include <iostream>
#include <algorithm>
using namespace std;
int a[100005];
int main()
{int n,k;cin>>n>>k;for(int i=1;i<=n;i++){cin>>a[i];}nth_element(a+1,a+k,a+1+n);cout<<a[k]<<endl;return 0;
}

归并排序 

归并排序

#include <iostream>
using namespace std;
const int N = 100010;
int n;
int q[N],tmp[N];
void merge_sort(int q[],int l,int r)
{if(l>=r){return;}int mid=(l+r)>>1;merge_sort(q,l,mid);merge_sort(q,mid+1,r);int k=1,i=l,j=mid+1;while(i<=mid&&j<=r){if(q[i]<=q[j]){tmp[k++]=q[i++];}else{tmp[k++]=q[j++];}}while(i<=mid){tmp[k++]=q[i++];}while(j<=r){tmp[k++]=q[j++];}for(int i=l,j=1;i<=r;i++,j++){q[i]=tmp[j];}
}
int main()
{scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&q[i]);}merge_sort(q,1,n);for(int i=1;i<=n;i++){printf("%d ",q[i]);}return 0;
}

逆序对的数量

#include <iostream>
using namespace std;
typedef long long ll;
const int N = 100010;
int n;
int q[N],tmp[N];
ll merge_sort(int l,int r)
{if(l>=r){return 0;}int mid = (l+r)>>1;ll res=merge_sort(l,mid)+merge_sort(mid+1,r);//归并的过程int k=1,i=l,j=mid+1;while(i<=mid&&j<=r){if(q[i]<=q[j]){tmp[k++]=q[i++];}else{tmp[k++]=q[j++];res+=(mid-i+1);}}//扫尾while(i<=mid){tmp[k++]=q[i++];}while(j<=r){tmp[k++]=q[j++];}//物归原主for(int i=l,j=1;i<=r;i++,j++){q[i]=tmp[j];}return res;
}
int main()
{cin>>n;for(int i=1;i<=n;i++){cin>>q[i];}cout<<merge_sort(1,n)<<endl;return 0;
}

二分 

数的范围

#include <iostream>
using namespace std;
const int N = 100010;
int q[N];
int main()
{int n,m;cin>>n>>m;for(int i=0;i<n;i++){cin>>q[i];}while(m--){int x;cin>>x;int l=0,r=n-1;while(l<r){int mid=l+r>>1;if(q[mid]>=x){r=mid;}else l=mid+1;}if(q[l]!=x){cout<<"-1 -1"<<endl;}else{cout<<l<<" ";l=0;r=n-1;while(l<r){int mid=l+r+1>>1;if(q[mid]<=x){l=mid;}else r=mid-1;}cout<<l<<endl;}}return 0;
}

数的三次方根

#include <iostream>
using namespace std;
int main()
{double n;cin>>n;double l=-10000,r=10000;while(r-l>1e-8){double mid = (l+r)/2;if(mid*mid*mid>=n){r=mid;}else l=mid;}printf("%.6lf\n",l);return 0;
}

高精度

 高精度加法

Python一行就可以解决

print(int(input())+int(input()))

#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<int> add(vector<int> &A,vector<int> &B)
{vector<int>C;int t=0;for(int i=0;i<A.size()||i<B.size();i++){if(i<A.size()){t+=A[i];}if(i<B.size()){t+=B[i];}C.push_back(t%10);t/=10;}if(t){C.push_back(t);}return C;
}
int main()
{string a,b;cin>>a>>b;vector<int>A,B;for(int i=a.size()-1;i>=0;i--){A.push_back(a[i]-'0');}for(int i=b.size()-1;i>=0;i--){B.push_back(b[i]-'0');}auto C = add(A,B);for(int i=C.size()-1;i>=0;i--){cout<<C[i];}return 0;
}

高精度减法

#include <iostream>
#include <string>
#include <vector>
using namespace std;
//判断是否有A>=B
bool cmp(vector<int> &A,vector<int> &B)
{if(A.size()!=B.size()){return A.size()>B.size();}for(int i=A.size()-1;i>=0;i--){if(A[i]!=B[i]){return A[i]>B[i];}}return true;
}
//C=A-B
vector<int> sub(vector<int> &A,vector<int> &B)
{vector<int>C;int t=0;for(int i=0;i<A.size();i++){t=A[i]-t;if(i<B.size()){t-=B[i];}C.push_back((t+10)%10);if(t<0){t=1;}else t=0;}while(C.size()>1&&C.back()==0){C.pop_back();}return C;
}
int main()
{string a,b;cin>>a>>b;vector<int>A,B;for(int i=a.size()-1;i>=0;i--){A.push_back(a[i]-'0');}for(int i=b.size()-1;i>=0;i--){B.push_back(b[i]-'0');}if(cmp(A,B)){auto C = sub(A,B);for(int i=C.size()-1;i>=0;i--){cout<<C[i];}}else{auto C = sub(B,A);cout<<"-";for(int i=C.size()-1;i>=0;i--){cout<<C[i];}}return 0;
}

高精度乘法

#include <iostream>
#include <vector>
using namespace std;
vector<int> mul(vector<int>& A, int b)
{vector<int>C;int t = 0;for (int i = 0; i < A.size() || t; i++){if (i < A.size()){t += A[i] * b;}C.push_back(t % 10);t /= 10;}while(C.size()>1&&C.back()==0){C.pop_back();}return C;
}
int main()
{string a;int b;cin >> a >> b;vector<int>A;for (int i = a.size() - 1; i >= 0; i--){A.push_back(a[i] - '0');}auto C = mul(A, b);for (int i = C.size() - 1; i >= 0; i--){cout << C[i];}return 0;
}

高精度除法

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
vector<int> div(vector<int>& A, int b,int &r)
{vector<int>C;r=0;for (int i = A.size()-1; i >= 0; i--){r = r * 10 + A[i];C.push_back(r/b);r%=b;}reverse(C.begin(),C.end());while(C.size()>1&&C.back()==0){C.pop_back();}return C;
}
int main()
{string a;int b;cin >> a >> b;vector<int>A;for (int i = a.size() - 1; i >= 0; i--){A.push_back(a[i] - '0');}int r;auto C = div(A, b,r);for (int i = C.size() - 1; i >= 0; i--){cout << C[i];}cout<<endl<<r<<endl;return 0;
}

前缀和与差分

前缀和

#include <iostream>
using namespace std;
const int N = 100005;
int a[N],s[N];
int main()
{int n,m;cin>>n>>m;for(int i=1;i<=n;i++){cin>>a[i];}for(int i=1;i<=n;i++){s[i]=s[i-1]+a[i];}while(m--){int l,r;cin>>l>>r;cout<<s[r]-s[l-1]<<endl;}return 0;
}

子矩阵的和

#include <iostream>
using namespace std;
const int N = 1010;
int a[N][N], s[N][N];
int main() 
{int n, m, q;cin >> n >> m >> q;for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++){scanf("%d", &a[i][j]);s[i][j] = s[i][j - 1] + s[i - 1][j] - s[i - 1][j - 1] + a[i][j]; // 求前缀和}while (q--) {int x1,y1,x2,y2;scanf("%d%d%d%d", &x1, &y1, &x2, &y2);// 算子矩阵的和printf("%d\n", s[x2][y2] - s[x2][y1 - 1] - s[x1 - 1][y2] + s[x1 - 1][y1 - 1]); }return 0;
}

差分

#include <iostream>
using namespace std;
const int N = 100010;
int n,m;
int a[N],b[N];
void insert(int l,int r,int c)
{b[l]+=c;b[r+1]-=c;
}
int main()
{cin>>n>>m;for(int i=1;i<=n;i++){cin>>a[i];}for(int i=1;i<=n;i++){insert(i,i,a[i]);}while(m--){int l,r,c;cin>>l>>r>>c;insert(l,r,c);}for(int i=1;i<=n;i++){b[i]+=b[i-1];}for(int i=1;i<=n;i++){cout<<b[i]<<" ";}return 0;
}

差分矩阵

#include <iostream>
using namespace std;
const int N = 1010;
int n,m,q;
int a[N][N],b[N][N];
void insert(int x1,int y1,int x2,int y2,int c)
{b[x1][y1]+=c;b[x2+1][y1]-=c;b[x1][y2+1]-=c;b[x2+1][y2+1]+=c;
}
int main()
{cin>>n>>m>>q;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>a[i][j];}}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){insert(i,j,i,j,a[i][j]);}}while(q--){int x1,y1,x2,y2,c;cin>>x1>>y1>>x2>>y2>>c;insert(x1,y1,x2,y2,c);}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){b[i][j]+=b[i-1][j]+b[i][j-1]-b[i-1][j-1];}}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cout<<b[i][j]<<" ";}cout<<endl;}return 0;
}

双指针算法

最长连续不重复子序列

#include <iostream>
using namespace std;
const int N = 100010;
int n;
int a[N],s[N];
int main()
{cin>>n;for(int i=1;i<=n;i++){cin>>a[i];}int res=0;for(int i=1,j=1;i<=n;i++){s[a[i]]++;while(s[a[i]]>1){s[a[j]]--;j++;}res=max(res,i-j+1);}cout<<res<<endl;return 0;
}

数组元素的目标和

#include <iostream>
using namespace std;
const int N = 100005;
int n,m,x;
int a[N],b[N];
int main()
{cin>>n>>m>>x;for(int i=1;i<=n;i++){cin>>a[i];}for(int i=1;i<=m;i++){cin>>b[i];}for(int i=1,j=m;i<=n;i++){while(j>=1&&a[i]+b[j]>x) j--;if(a[i]+b[j]==x){cout<<i-1<<" "<<j-1<<endl;break;}}return 0;
}

判断子序列

#include <iostream>
using namespace std;
const int N = 100005;
int n,m;
int a[N],b[N];
int main()
{cin>>n>>m;for(int i=0;i<n;i++){cin>>a[i];}for(int i=0;i<m;i++){cin>>b[i];}int i=0,j=0;while(i<n&&j<m){if(a[i]==b[j])i++;j++;}if(i==n){cout<<"Yes"<<endl;}else{cout<<"No"<<endl;}return 0;
}

位运算

二进制中1的个数

#include <iostream>
using namespace std;
int lowbit(int x)
{return x & -x;
}
int main()
{int n;cin>>n;while(n--){int x;cin>>x;int res=0;while(x){x-=lowbit(x);res++;}cout<<res<<" ";}return 0;
}

离散化

区间和

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 300010; //n次插入和m次查询相关数据量的上界
int n, m;
int a[N];//存储坐标插入的值
int s[N];//存储数组a的前缀和
vector<int> alls;  //存储（所有与插入和查询有关的）坐标
vector<pair<int, int>> add, query; //存储插入和询问操作的数据int find(int x) 
{ //返回的是输入的坐标的离散化下标int l = 0, r = alls.size() - 1;while (l < r) {int mid = l + r >> 1;if (alls[mid] >= x) r = mid;else l = mid + 1;}return r + 1;
}int main() 
{scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++) {int x, c;scanf("%d%d", &x, &c);add.push_back({x, c});alls.push_back(x);}for (int i = 1; i <= m; i++) {int l , r;scanf("%d%d", &l, &r);query.push_back({l, r});alls.push_back(l);alls.push_back(r);}//排序，去重sort(alls.begin(), alls.end());alls.erase(unique(alls.begin(), alls.end()), alls.end());//执行前n次插入操作for (auto item : add) {int x = find(item.first);a[x] += item.second;}//前缀和for (int i = 1; i <= alls.size(); i++) s[i] = s[i-1] + a[i];//处理后m次询问操作for (auto item : query) {int l = find(item.first);int r = find(item.second);printf("%d\n", s[r] - s[l-1]);}return 0;
}

区间和并

区间和并

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
typedef pair<int,int>pii;
const int N = 100010;
int n;
vector<pii>segs;
void merge(vector<pii>&segs)
{vector<pii>res;sort(segs.begin(),segs.end());int st=-2e9,ed=-2e9;for(auto seg:segs){if(ed<seg.first){if(ed!=-2e9) res.push_back({st,ed});st=seg.first;ed=seg.second;}else ed=max(ed,seg.second);}if(st!=2e9) res.push_back({st,ed});cout<<res.size()<<endl;
}
int main()
{cin>>n;for(int i=0;i<n;i++){int l,r;cin>>l>>r;segs.push_back({l,r});}merge(segs);return 0;
}