UVA 12670 数位DP

UVA 12670
题目链接：
https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4536
题意：
给一个数字范围[x,y]，问在这个范围内所有数表示成二进制后，“1”的个数和。
思路：
简单的数位DP，用那种BIT的思想去理解就好。
源码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <string>
using namespace std;
#define LL long long
const int MAXN = 70;
LL dp[MAXN];
LL C[MAXN];
LL cal(LL a)
{
    string str = "";
    while(a){
        str += a % 2 + '0';
        a /= 2;
    }
    LL ans = 0;
// cout << " str = " << str<<endl;
// str.reverse();
    int len = str.length();
    int cnt = 0;
    for(int i = len - 1 ; i >= 0 ; i--){
        if(str[i] == '1'){
// printf("i = %d, dp[i] = %lld\n", i, dp[i]);
            ans += dp[i] + cnt * C[i];
            cnt++;
        }
    }
    ans += cnt;
    return ans;
}
void init()
{
    dp[0] = 0;
    C[0] = 1;
    for(int i = 1 ; i < MAXN ; i++){
        dp[i] = dp[i - 1] * 2 + C[i - 1];
        C[i] = C[i - 1] * 2;
    }
}
int main()
{
    init();
    LL a, b;
    while(scanf("%lld%lld", &a, &b) != EOF){
        LL ans = cal(b) - cal(a - 1);
// printf("cal1 = %lld, cal2 = %lld\n", cal(b), cal(a - 1));
        printf("%lld\n", ans);
    }
    return 0;
}