CF103 D. Time to Raid Cowavans
阿新 • • 發佈:2021-06-30
題目傳送門:https://codeforces.com/problemset/problem/103/D
題目大意:
給你一串長度為\(n\)的序列\(A\),每次詢問給一組\((a,b)\),求\(\sum\limits_{i=0}^{\lfloor\frac{n-a}{b}\rfloor}A_{a+bi}\)
考慮離線,對所有詢問的\(b\)進行排序
然後依據\(b\)與\(\sqrt n\)的大小關係,分為兩種處理方法
- \(b<\sqrt n\),對序列\(A\)求跨步為\(b\)的字尾和\(S\),直接輸出\(S_a\)即可
- \(b>\sqrt n\),每次詢問從\(a\)開始暴力向後跳,統計答案
時間複雜度分析:
- \(b<\sqrt n\),\(b\)僅有\(\sqrt n\)種取值,每次取值需要\(O(n)\)統計字尾和,輸出同\(b\)的詢問僅需\(O(1)\),故該部分時間複雜度為\(O(n\sqrt n)\)
- \(b>\sqrt n\),\(b\)的取值數為\(O(n)\);但此時,每次暴力僅需要\(O(\sqrt n)\)的次數就可以統計完畢,故該部分時間複雜度為\(O(n\sqrt n)\)
綜上所述,總時間複雜度為\(O(n\sqrt n)\)
/*program from Wolfycz*/ #include<map> #include<cmath> #include<cstdio> #include<vector> #include<cstring> #include<iostream> #include<algorithm> #define Fi first #define Se second #define ll_inf 1e18 #define MK make_pair #define sqr(x) ((x)*(x)) #define pii pair<int,int> #define int_inf 0x7f7f7f7f using namespace std; typedef long long ll; typedef unsigned int ui; typedef unsigned long long ull; inline char gc(){ static char buf[1000000],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++; } template<typename T>inline T frd(T x){ int f=1; char ch=gc(); for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } template<typename T>inline T read(T x){ int f=1; char ch=getchar(); for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } inline void print(int x){ if (x<0) putchar('-'),x=-x; if (x>9) print(x/10); putchar(x%10+'0'); } const int N=3e5; int A[N+10]; ll S[N+10],Ans[N+10]; struct opt{ int a,b,ID; opt(int _a=0,int _b=0,int _ID=0){a=_a,b=_b,ID=_ID;} bool operator <(const opt &tis)const{return b<tis.b;} }B[N+10]; int main(){ // freopen("data.in","r",stdin); // freopen("ycz.out","w",stdout); int n=read(0),len=sqrt(n); for (int i=0;i<n;i++) A[i]=read(0); int m=read(0); for (int i=1;i<=m;i++){ int a=read(0)-1,b=read(0); B[i]=opt(a,b,i); } sort(B+1,B+1+m); for (int i=1;i<=m;i++){ if (B[i].b>=len){ for (int j=B[i].a;j<n;j+=B[i].b) Ans[B[i].ID]+=A[j]; continue; } if (B[i].b!=B[i-1].b){ for (int j=n-1;~j;j--){ S[j]=A[j]; if (j+B[i].b<n) S[j]+=S[j+B[i].b]; } } Ans[B[i].ID]=S[B[i].a]; } for (int i=1;i<=m;i++) printf("%lld\n",Ans[i]); return 0; }