23 solutions
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1
这种题目,必须折磨亿下~~~
1.线段树写法
#include<bits/stdc++.h> typedef long long ll; using namespace std; struct node{ int l,r; ll num; }tree[100001]; ll a[100001]={},n,tag[100001]; void ctf(ll c){ tree[c].num=tree[c<<1].num+tree[(c<<1)+1].num; } void createTree(ll sp,ll l,ll r){ tree[sp].l=l; tree[sp].r=r; if(l==r){ tree[sp].num=a[l]; return; } ll mid=(l+r)>>1; createTree(sp<<1,l,mid); createTree((sp<<1)+1,mid+1,r); ctf(sp); } void cltag(ll sp){ ll mid=(tree[sp].l+tree[sp].r)>>1; if(tag[sp]){ tree[sp<<1].num+=tag[sp]*(mid-tree[sp].l+1); tree[(sp<<1)+1].num+=tag[sp]*(tree[sp].r-mid); tag[sp<<1]+=tag[sp]; tag[(sp<<1)+1]+=tag[sp]; tag[sp]=0; } } void pointChange(ll sp,ll pt,ll nn){ if(tree[sp].l==tree[sp].r){ tree[sp].num+=nn; return; } ll mid=(tree[sp].l+tree[sp].r)>>1; if(pt<=mid)pointChange(sp<<1,pt,nn); else pointChange((sp<<1)+1,pt,nn); ctf(sp); } void areaChange(ll sp,ll l,ll r,ll c){ if(tree[sp].l>=l&&tree[sp].r<=r){ tree[sp].num+=(tree[sp].r-tree[sp].l+1)*c; tag[sp]+=c; return; } cltag(sp); ll mid=(tree[sp].l+tree[sp].r)>>1; if(l<=mid){ areaChange(sp<<1,l,r,c); } if(mid<r){ areaChange((sp<<1)+1,l,r,c); } ctf(sp); } ll areaAsk(ll sp,ll l,ll r){ if(tree[sp].l>=l&&tree[sp].r<=r)return tree[sp].num; cltag(sp); ll mid=(tree[sp].l+tree[sp].r)>>1,maxn=0; if(l<=mid){ maxn=maxn+areaAsk(sp<<1,l,r); } if(mid<r){ maxn=maxn+areaAsk((sp<<1)+1,l,r); } return maxn; } int main(){ ll T; n=2;T=1; for(ll i=1;i<=2;i++)cin>>a[i]; createTree(1,1,2); cout<<areaAsk(1,1,2); }
2.树状数组写法
#include<bits/stdc++.h> typedef long long ll; using namespace std; ll lowbit(ll __n){return __n&-__n;} namespace ACAQ{ ll a[1000001]; ll trb[1000001]; ll trib[1000001]; ll n; void changeArea(ll l,ll r,ll adn){ r++; for(ll i=l;i<=n;i+=lowbit(i)){ trb[i]+=adn; trib[i]+=adn*l; } for(ll i=r;i<=n;i+=lowbit(i)){ trb[i]-=adn; trib[i]-=adn*r; } return; } ll getAreaSum(ll l,ll r){ ll sum1=0,sum2=0; l--; for(ll i=l;i;i-=lowbit(i))sum1+=trb[i]*(l+1)-trib[i]; for(ll i=r;i;i-=lowbit(i))sum2+=trb[i]*(r+1)-trib[i]; return sum2-sum1; } void read(){ for(ll i=1;i<=n;i++){ cin>>a[i]; for(ll j=i;j<=n;j+=lowbit(j)){ trb[j]+=a[i]-a[i-1]; trib[j]+=(a[i]-a[i-1])*i; } } } //-----example-----// void example(){ ll t;n=2,t=1; read(); while(t--){ cout<<getAreaSum(1,2); } } } namespace ACPQ{ } namespace PCAQ{ } int main(){ ACAQ::example(); }
3.分块写法
#include<bits/stdc++.h> using namespace std; typedef long long ll; ll arr[1000001]; ll pos[1000001]; ll tag[1000001]; ll blk[1000001]; ll cfs[1000001][2]; ll n; void blockInit(){ ll sn=sqrt(n),k=n/sn+bool(n%sn); for(ll i=1;i<=k;i++){ cfs[i][0]=sn*(i-1)+1; cfs[i][1]=sn*i; } cfs[k][1]=n; for(ll i=1;i<=k;i++){ for(ll j=cfs[i][0];j<=cfs[i][1];j++){ pos[j]=i;blk[i]+=arr[j]; } } } void areaAdd(ll l,ll r,ll x){ ll lx=pos[l],rx=pos[r]; if(lx==rx){ for(ll i=l;i<=r;i++)arr[i]+=x; blk[lx]+=(r-l+1)*x; return; } for(ll i=lx+1;i<rx;i++)tag[i]+=x; for(ll i=l;i<=cfs[lx][1];i++)arr[i]+=x; blk[lx]+=(cfs[lx][1]-l+1)*x; for(ll i=cfs[rx][0];i<=r;i++)arr[i]+=x; blk[rx]+=(r-cfs[rx][0]+1)*x; } ll getAreaSum(ll l,ll r){ ll sum=0; ll lx=pos[l],rx=pos[r]; if(lx==rx){ for(ll i=l;i<=r;i++)sum+=arr[i]; return sum+tag[lx]*(r-l+1); } for(ll i=lx+1;i<rx;i++)sum+=blk[i]+tag[i]*(cfs[i][1]-cfs[i][0]+1); for(ll i=l;i<=cfs[lx][1];i++)sum+=arr[i]; sum+=tag[lx]*(cfs[lx][1]-l+1); for(ll i=cfs[rx][0];i<=r;i++)sum+=arr[i]; sum+=tag[rx]*(r-cfs[rx][0]+1); return sum; } int main(){ ll m; n=2,m=1; for(ll i=1;i<=n;i++)cin>>arr[i]; blockInit(); // for(ll i=1;i<=n;i++)cout<<cfs[i][0]<<" "<<cfs[i][1]<<" "<<pos[i]<<endl; for(ll i=1;i<=m;i++){ cout<<getAreaSum(1,2); } }
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1
#include <bits/stdc++.h> #define int long long using namespace std; struct str{ int l,r,v,add; }tr[400005]; int a[100005]; void pushup(int p) { tr[p].v=tr[p<<1].v+tr[p<<1|1].v; } void pushdown(int p) { int mid=(tr[p].l+tr[p].r)>>1; if(tr[p].add!=0) { tr[p<<1].v+=tr[p].add*(mid-tr[p].l+1); tr[p<<1|1].v+=tr[p].add*(tr[p].r-mid); tr[p<<1].add+=tr[p].add; tr[p<<1|1].add+=tr[p].add; tr[p].add=0; } } void build(int p,int x,int y) { tr[p].l=x; tr[p].r=y; if(x==y) { tr[p].v=a[x]; return; } int mid=(x+y)>>1; build(p<<1,x,mid); build(p<<1|1,mid+1,y); pushup(p); } void update(int p,int x,int y,int v) { if (x<=tr[p].l&&tr[p].r<=y) { tr[p].v+=v*(tr[p].r-tr[p].l+1); tr[p].add+=v; return; } pushdown(p); int mid=(tr[p].l+tr[p].r)>>1; if(x<=mid){ update(p<<1,x,y,v); } if(y>mid){ update(p<<1|1,x,y,v); } pushup(p); } int query(int p,int x,int y) { if (x<=tr[p].l&&tr[p].r<=y){ return tr[p].v; } pushdown(p); int mid=(tr[p].l+tr[p].r)>>1; int ans=0; if (x<=mid){ ans+=query(p<<1,x,y); } if(y>mid){ ans+=query(p<<1|1,x,y); } return ans; } signed main() { ios::sync_with_stdio(false); cin.tie(0),cout.tie(0); for(int i=1;i<=2;i++){ cin>>a[i]; } build(1,1,2); cout<<query(1,1,2); }
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#include <bits/stdc++.h> #define jiajia ++ #define jiadengyu += #define jia + #define jian - #define cheng * #define chu / #define shuchu cout #define fanhui return #define false true #define true false #define dingyi int #define shuru cin #define xiaoyudengyu <= #define dengyu = #define dingyipanduanleizifuchuang bool #define kaishi { #define jieshu } #define qiankuohao ( #define fenhao ; #define shurufuhao >> #define shuchufuhao << #define ling 0 #define yi 1 #define xunhuan for #define douhao , #define zheng true #define ruguo if #define houkuohao ) #define shiyong using #define mingmingkongjian namespace shiyong mingmingkongjian std fenhao dingyipanduanleizifuchuang o qiankuohao dingyi p houkuohao kaishi fanhui zheng fenhao jieshu dingyi main qiankuohao houkuohao kaishi dingyi a douhao b fenhao shuru shurufuhao a shurufuhao b fenhao shuchu shuchufuhao a jia b fenhao fanhui ling fenhao jieshu //抽象代码
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-6
#include<bits/stdc++.h> #define ll long long using namespace std; unsigned ll a,b; unsigned ll c; int main(){ scanf("%d%d",&a,&b); c=max(a,min(a+b,max(a+b,max(a+b+b,max(a+b+b+b,max(a+b+b+b+b,a)))))); /*原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动! 原神,启动!*/ printf("%d",c); return 0; }
看看这个:A+B问题进阶版
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-7
#include <set> #include <ios> #include <list> #include <cmath> #include <ctime> #include <queue> #include <deque> #include <stack> #include <vector> #include <bitset> #include <cctype> #include <cerrno> #include <cwchar> #include <cstdio> #include <fenv.h> #include <iosfwd> #include <string> #include <limits> #include <math.h> #include <cstdlib> #include <iomanip> #include <clocale> #include <complex> #include <cstring> #include <cstring> #include <cwctype> #include <istream> #include <ostream> #include <sstream> #include <fstream> #include <utility> #include <stdio.h> #include <iostream> #include <stdint.h> #include <string.h> #include <tgmath.h> #include <complex.h> #include <algorithm> #include <exception> #include <stdbool.h> #include <stdexcept> #include <streambuf> #include <functional> #include <inttypes.h> #include <bits/stdc++.h> using namespace std; int k(int x){ if(x == 0) return 1; if(x == 1) x = 1; return k(x - 1); } int main(){ int a,b,c; cin >> a >> b; for(int i = 1;i < 1;i++){ if(1 + 1 == 2) c = a + b; else c = a + b; for(int j = 1;j < 1;j++){ if(1 + 1 == 2) c = a + b; else c = a + b; for(int k = 1;k < 1;k++){ if(1 + 1 == 2) c = a + b; else c = a + b; for(int q = 1;q < 1;q++){ if(1 + 1 == 2) c = a + b; else c = a + b; for(int w = 1;w < 1;w++){ if(1 + 1 == 2) c = a + b; else c = a + b; for(int e = 1;e < 1;e++){ if(1 + 1 == 2) c = a + b; else c = a + b; for(int r = 1;r < 1;i++){ if(1 + 1 == 2) c = a + b; else c = a + b; for(int y = 1;y < 1;y++){ if(1 + 1 == 2) c = a + b; else c = a + b; while(1) break; while(1) break; while(1) break; while(1) break; while(1) break; } while(1) break; while(1) break; while(1) break; while(1) break; while(1) break; } while(1) break; while(1) break; while(1) break; while(1) break; while(1) break; } while(1) break; while(1) break; while(1) break; while(1) break; while(1) break; } while(1) break; while(1) break; while(1) break; while(1) break; while(1) break; } while(1) break; while(1) break; while(1) break; while(1) break; while(1) break; } while(1) break; while(1) break; while(1) break; while(1) break; while(1) break; } while(1) break; while(1) break; while(1) break; while(1) break; while(1) break; } while(1) break; while(1) break; while(1) break; while(1) break; while(1) break; //cout << endl; //cout << endl; //cout << endl; //cout << endl; //cout << endl; //cout << endl; //cout << endl; //cout << endl; //cout << endl; //cout << endl; if(k(k(k(k(k(k(k(k(k(k(k(k(k(k(k(k(114514))))))))))))))))){ k(1); } c = a + b; cout << c; return 0; }
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-7
#include <iostream> using namespace std; int main(){ int a , b; cin >> a >> b; int j; j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = 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a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b;j = a + b; j = a + b; cout << j; return 0; }
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-10
#include<bits/stdc++.h> using namespace std; int z1(int s){ int q=0; for(int i=2;i<s;i++){ if(s%i==0) q=1; } if(q==1) return 0; else return 1; } void z(long long a,long long b){ long long A,A1,B1,B,c,o=1,h=1; for(int i=1;i<=a;i++){ if(z1(i)&&a%i==0){ A=a/i;A1=i;o*=o; } } for(int i=1;i<=b;i++){ if(z1(i)&&b%i==0){ B=b/i;B1=i; } } c=(A*A1+B*B1)*o; cout<<c/h; } int main(){ long long a_,b_; cin>>a_>>b_; z(a_,b_); return 0; }
Information
- ID
- 521
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 5
- Tags
- # Submissions
- 402
- Accepted
- 158
- Uploaded By