求 ∑i=1naimod1000000007$\sum _{i=1}^n{a}_i\mathrm{mod}1000000007$

There are multiple test cases. The first line of input contains an integer T(1≤T≤105)$T(1\le T\le {10}^5)$, indicating the number of test cases. For each test case:
The first line contains an integer n(1≤n≤1018)$n(1\le n\le {10}^{18})$.

思路

an$a_n$ 序列中某個值 x$x$ 出現的次數是 max{k+1|2k整除x}$max\{k+1|2^k整除x\}$

然後二分得到 an$a_n$

然後可以 logan$\log a_n$ 時間內求出

然後 減掉多加的部分就得到了答案

#pragma GCC optimize ("O2")
#include<bits/stdc++.h>
 

using namespace std;

const int MAXN = (int)1e5 + 5;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-8;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const int MOD = 1000000007;

ll d[64];

ll f(ll n) {
    ll num = 0;
    for(ll i = 1; d[i] <= n; ++i) {
        num += i * ((n / d[i] + 1
 
) / 2);
    }
    return num + 1;
}

int sum(ll n) {
    ll res = 0;
    for(ll i = 1; d[i] <= n; ++i) {
        //易錯
        (res += i * ((d[i] % MOD + d[i] % MOD + (2 * d[i]) % MOD * (((n / d[i] + 1) / 2 - 1) % MOD) % MOD) % MOD * ((n / d[i] + 1) / 2 % MOD) % MOD * 500000004 % MOD) % MOD) %= MOD;
    }
    return (int)res;
}

ll work(ll n) {
    // 一定要加速收斂
    ll low = max(n / 2, 0LL), top = n / 2 + 30;
    while(top - low > 1) {
        ll mid = (low + top) >> 1;
        if(f(mid) >= n) top = mid;
        else low = mid;
    }
    return top;
}

void slove(ll n) {
    ll sop = work(n);
    ll pos = f(sop);
    int ans = sum(sop) + 1;
    ans -= (pos - n) * sop % MOD;
    printf("%d\n", (ans + MOD) % MOD);
}

ll read(ll &x) {
    char c;
    while((c = getchar()) <= 32);
    for(x = 0; c >= '0'; c = getchar()) x = x * 10 + c - '0';
}

int main()
{
    d[1] = 1;
    for(int i = 2; i < 64; ++i) d[i] = d[i - 1] << 1;
    int T;ll n;
    while(scanf("%d", &T) != EOF) {
        while(T--) {
            read(n);
            slove(n);
        }
    }
    return 0;
}