Codeforces Beta Round #91 (Div. 1 Only) A. luckSum(区间分块求解)

这道题的主要解法在对1~100000000 中的luckNumber进行分块求和;

例如, luckSum[0] = Sum[1,4]  luckSum[1] = Sum[5,7] luckSum[2] = Sum[8,44] ......................

然后根据l,r 的区间跨度来求和;

所在区间用一个int findUnder(int l)来求取;

#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;

typedef long long ll;

const int maxn = 1e9;
ll luckNumber[1500][2];
ll luckSum[1500];

void init(){
    int cnt=2;
    long long num;
    queue<long long> que; 
    
    luckNumber[0][0]=1;
    luckNumber[0][1]=4;
    luckNumber[1][0]=5;
    luckNumber[1][1]=7;
    
    que.push(4);
    que.push(7);
    
    while(!que.empty()){
        num = que.front();
        que.pop();
        
        if(num*10+4<=maxn) luckNumber[cnt][0] = luckNumber[cnt-1][1]+1,luckNumber[cnt++][1] = num*10+4,que.push(num*10+4);
        else break;
        
        if(num*10+7<=maxn) luckNumber[cnt][0] = luckNumber[cnt-1][1]+1,luckNumber[cnt++][1] = num*10+7,que.push(num*10+7);
        else break;
        
    }
    luckNumber[cnt][0]=luckNumber[cnt-1][1]+1;
    luckNumber[cnt][1]=4444444444;
    
    for(int i=0;i<cnt;i++){
        luckSum[i] = (luckNumber[i][1]-luckNumber[i][0]+1)*luckNumber[i][1];
        //cout<<luckSum[i]<<endl;
    }
    //cout<<cnt<<endl;
        
}

int findUnder(int l){
    
    for(int i=0;i<=1022;i++){
        if(l<=luckNumber[i][1]&&l>=luckNumber[i][0]) return i;
    }
    
}

int main(){
    init(); 
    int l,r,lSize,rSize;
    ll sum;
    scanf("%d%d",&l,&r);
    
    lSize = findUnder(l);
    rSize = findUnder(r);
    sum = 0;
    
    sum  = (min(luckNumber[lSize][1],(ll)r)-l+1)*luckNumber[lSize][1];
    for(int i=lSize+1;i<rSize;i++) sum+=luckSum[i];
    if(lSize!=rSize){
        sum += (r-luckNumber[rSize][0]+1)*luckNumber[rSize][1];
    }
    
    printf("%I64d
",sum);
    
    return 0;
} 

由cnt可以知道每次循环至多需要2^9次, 而至多只有5层循环 ，O(5*2^9);