【預先處理】Codeforces Beta Round #80 (Div. 1 Only) ——Time to Raid Cowavans

這題的官方時限是4s，開始直接爆，果然TLE了==後來最佳化了一下，1420ms過了。。。所以說坑爹啊！！！

最佳化方法是因為題目中的b是數與數之間的間隔，當間隔大於600時可以直接遍曆，當小於600時就要記錄處於當前間隔的情況有多少種，然後從後向前推，推得的tmp[j]表示j~n間隔為i的和（相當於預先處理），這樣就把最壞的複雜度降了下來，以後就直接O(1)就找到當前的答案。

#include <map>#include <set>#include <list>#include <queue>#include <deque>#include <stack>#include <string>#include <time.h>#include <cstdio>#include <math.h>#include <iomanip>#include <cstdlib>#include <limits.h>#include <string.h>#include <iostream>#include <fstream>#include <algorithm>using namespace std;#define LL long long#define MIN INT_MIN#define MAX INT_MAX#define PI acos(-1.0)#define FRE freopen ("input.txt","r",stdin)#define FF  freopen ("output.txt","w",stdout)#define eps 1e-8#define N 300005LL num[N];LL tmp[N];struct node {    int id;    int a,b;    LL ans;    node(){        ans = 0;    }}p[N];vector<node> v[N];int main(){    int n;    scanf("%d",&n);    int i,j;    for (i = 1; i <= n; i++) {        scanf("%d",&num[i]);    }    int m;    scanf("%d",&m);    for (i = 1; i <= m; i++) {        int a,b;        scanf("%d%d",&a,&b);        p[i].a = a;        p[i].b = b;        p[i].id = i;        p[i].ans = 0;        if (b > 400) {            for (j = a; j <= n; j+=b ){                p[i].ans += num[j];            }        }else        v[b].push_back(p[i]);    }    for (i = 1; i <= 400; i++ ){        if (v[i].size() > 0){            for (j = n; j >= 1; j--) {                tmp[j] = num[j];                if (i+j <= n) {                    tmp[j] += tmp[i+j];                }            }            for (j = 0; j < v[i].size(); j++) {                p[v[i][j].id].ans = tmp[v[i][j].a];            }        }    }    for (i = 1; i <= m; i++){        printf("%I64d\n",p[i].ans);    }    return 0;}