Binary Indexed Tree HDU-5921 (贡献，计数，数位DP)

Binary Indexed Tree HDU-5921 (贡献，计数)

题目链接：

HDU - 5921

题意：

给定你一个整数\(n\in[1,10^{18}]\)，问你\(ans=\sum_{i=1}^{n}\sum_{j=1}^{i-1}cost(j,i)\)

其中\(cost(j,i)\)是 树状数组区间修改时\(add(l-1,-val),add(r,val)\)。实际被更改的位置个数。

思路：

通过分析可以发现\(cost(j,i)\)等于\(g(i)+g(j-1)-2\times g(lcp(i,j-1))\)

其中：

\(g(x)\)为\(\mathit x\) 在二进制表示法中\(\text 1\)的个数。

\(lcp(i,j)\)是\(i,j\)在二进制表示法中补上前导零使其等长度后的最长公共前缀。

那么答案

\[ans=\frac{\sum_{i=0}^{n}\sum_{j=0}^{n}(g(i)+g(j)-2\times g(lcp(i,j)))}{2} \\ =\frac{(n+1)\sum_{i=0}^{n}(2\times g(i))-2\times \sum_{i=0}^{n}\sum_{j=0}^{n}g(lcp(i,j))}{2} \\ =(n+1)\sum_{i=0}^{n}\times g(i)-\sum_{i=0}^{n}\sum_{j=0}^{n}g(lcp(i,j)) \]

然后来思考如何计算\(g(x),g(lcp(i,j))\)，

我们需要两个额外的数组来辅助计算：

\(r[i]\)代表将n二进制拆位后第\(\mathit i\)为右边的数位组成的数最大值。

\(l[i]\)代表将n二进制拆位后第\(\mathit i\)为左边的数位组成的数最大值。

计算\(g(x)\)分为两类贡献：

• 到第\(\mathit i\)位，左边的数位不改变，那么为了不大于原数，右边的数应取\([0,r[i-1]]\)。

即当\(w_i=1\)时，有\(r[i-1]+1\)的贡献，当\(w_i=0\)时，没有贡献。

• 到第\(\mathit i\)位，左边的数位改变了，左边的数应取\([0,l[i+1]-1]\).那么此时右边的数可以任意取值，有\(2^i\)种取值。因为高位已经小于上限了，该位置一定可以取1，所以有\(l[i+1]*2^i\)的贡献。

两两lcp的部分，则与之类似。

代码:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x)  if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int len;
ll w[65];
ll base[65], l[65], r[65];
const ll mod = 1e9 + 7ll;
ll getf()
{
    ll res = 0ll;
    for (int i = len; i >= 0; --i) {
        if (w[i]) {
            res++;
            res %= mod;
            if (i > 0) {
                res += r[i - 1];
                res %= mod;
            }
        }
        res += l[i + 1] * base[i] % mod;
        res %= mod;
    }
    return res;
}
ll getlcp()
{
    ll res = 0ll;
    for (int i = len; i >= 0; --i) {
        if (w[i]) {
            if (i > 0) {
                res += (r[i - 1] + 1) * (r[i - 1] + 1) % mod;
                res %= mod;
            } else {
                res += 1ll;
                res %= mod;
            }
        }
        res += l[i + 1] * base[i] % mod * base[i] % mod;
        res %= mod;
    }
    return res;
}

int main()
{
#if DEBUG_Switch
    freopen("D:\\code\\input.txt", "r", stdin);
#endif
    //freopen("D:\\code\\output.txt","w",stdout);
    int t;
    t = readint();
    int icase = 1;
    base[0] = 1ll;
    repd(i, 1, 64) {
        base[i] = base[i - 1] * 2ll % mod;
    }
    while (t--) {
        ll n = readll();
        ll temp = n % mod;
        len = 0;
        while (n) {
            w[len++] = n & 1;
            n >>= 1;
        }
        --len;
        l[len + 1] = 0ll;
        r[0] = w[0];
        for (int i = 1; i <= len; ++i) {
            r[i] = (r[i - 1] + w[i] * (1ll << i) % mod) % mod;
        }
        for (int i = len; i >= 0; --i) {
            l[i] = ((l[i + 1] << 1) % mod + w[i]) % mod;
        }
        ll ans = getf() * (temp + 1) % mod;
        ans = (ans - getlcp() + mod) % mod;
        printf("Case #%d: %lld\n", icase++, ans);
    }

    return 0;
}