题目链接
题意：给你给无限长的序列 A A A，让你求出 ∑ i = l r A i \sum_{i=l}^{r}A_i ∑i=lr​Ai​.
思路：转化一下，即求 ∑ i = 1 r A i − ∑ i = 1 l − 1 A i \sum_{i=1}^{r}A_i-\sum_{i=1}^{l-1}A_i ∑i=1r​Ai​−∑i=1l−1​Ai​。那么问题就好解决了，不断的倍增加上当前段的贡献，最后去掉多余的即可。

#include<bits/stdc++.h>

#define LL long long
#define fi first
#define se second
#define mp make_pair
#define pb push_back

using namespace std;

const int N = 2e5 + 11;
const LL mod = 1e9 + 7;
LL l, r, now = 1, sta, A, B;
LL odd, even;
LL get(LL g)
{
    A = B =  0;now = sta = 1;odd = even = 0;LL ans=0;
    while(1){
    	if(sta)ans+=(now/2)%mod*((B+now/2)%mod)%mod;
    	else ans+=(now/2)%mod*((A+now/2)%mod)%mod;
    	if(sta)B+=now;
    	else A+=now;
    	if(now>g)break;
    	now*=2;
    	sta^=1;
    }
    now--;
    sta^=1;
    if(sta)ans-=(now-g)%mod*((A-(now-g)+mod)%mod)%mod;
    else ans-=(now-g)%mod*((B-(now-g)+mod)%mod)%mod;
    return (ans+mod)%mod;
}
int main()
{
    ios::sync_with_stdio(false);
    cin >> l >> r;
    cout << (get(r) - get(l - 1) + mod) % mod << endl;
    return 0;
}