关于树状数组

树状数组是一种十分优秀的数据结构，拥有常数非常小的特点，好写好调，在一些应用上比线段树要优秀许多。下面我来介绍下树状数组（基础知识请看蓝书或其他神犇的blog，蒟蒻在这里就不多提了）。

区间修改&&区间查询

如题，已知一个数列，你需要进行下面两种操作：
1.将某区间每一个数数加上x
2.求出某一个数的和

这里要引入差分数组这种东西，我们记d[i] = a[i] - a[i-1]（a为原数组），这样我们记sigma（d[i]） = a[i] ，为什么呢，观察式子sigma(d[i]) = a[1] + a[2] - a[1] +ａ［３］．．．这样一直下去就得到了我们的原数组。
有什么用呢？如果我们往一段区间上加ｋ，在差分数组上如何体现呢？我们举个栗子：
ａ：１,２,３,４,５
ｄ：１,１,１,１,１
２～４加１
如果我们盲目的在２到４上加１，就会发现会影响后面的数（因为是前缀和），所以我们在２这个位置加一，用树状数组更新，在５的位置减一用树状数组更新就ｏｋ了
ａ：１，３，４，５，５
ｄ：１，２，１，１，０
这样就没问题了吧？总的时间是操作数×ｌｏｇ（ｎ）。

/*************************************************************************    > Author: Drinkwater-cnyali    > Created Time: 2017/6/13 10:22:21 ************************************************************************/#include<iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;#define REP(i, a, b) for(register int i = (a), i##_end_ = (b); i <= i##_end_; ++ i)#define DREP(i, a, b) for(register int i = (a), i##_end_ = (b); i >= i##_end_; -- i)char buff[1<<25], *buf = buff;#define mem(a, b) memset((a), b, sizeof(a))template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }int Fread(){    int sum = 0, fg = 1;    for(; !isdigit(*buf) ; ++buf)if(*buf == '-')fg = -1;    for(; isdigit(*buf); ++buf)sum = sum * 10 + *buf - '0';    return sum * fg;}int read(){    int sum = 0,fg = 1;char c = getchar();    while(c < '0' || c > '9') { if (c == '-') fg = -1; c = getchar(); }    while(c >= '0' && c <= '9') { sum = sum * 10 + c - '0'; c = getchar(); }    return sum * fg;}const int maxn = 1000000;const int inf = 0x3f3f3f3f;int n,m;int c[maxn],a[maxn];inline int lowbit(int x) {return (x & (-x));}inline void updata(int x,int num){    for(; x <= n; x += lowbit(x))        c[x] += num;}inline int query(int x){    int ans = 0;    for( ; x > 0; x -= lowbit(x))        ans += c[x];    return ans;}   int main(){    fread(buff,1,1<<25,stdin);    n = Fread(), m = Fread();    REP(i,1,n)a[i] = Fread(), updata(i,a[i] - a[i-1]);    REP(i,1,m)    {        int t = Fread();        if(t == 1)        {            int x = Fread(), y = Fread(),k = Fread();            updata(x,k),updata(y+1,-k);        }        if(t == 2)        {            int x = Fread();            printf("%d\n",query(x));        }       }       return 0;}

如题，已知一个数列，你需要进行下面两种操作：
1.将某区间每一个数加上x
2.求出某区间每一个数的和

这道题显然是线段树区间更新的裸题，这里我们要对式子进行变形
一个区间和我们记为sigma（ａ[i]） = d[1] + (d[1] + d[2]) + (d[1] + d[2]+ d[3]) + ..+(…d[n])
= nd[1] - (n-1)d[2] + ..+d[n]
= n * (d[1] + ..+d[n]) - (0d[1] + 1d[2] + ..+(n-1)d[n])
这样我们就只需要维护一个d的前缀和和(i-1)*d[i]的前缀和就好了。

/*************************************************************************    > Author: Drinkwater-cnyali    > Created Time: 2017/6/13 15:36:58 ************************************************************************/#prag\ma GCC optimize("O4")#include<iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;#define REP(i, a, b) for(register int i = (a), i##_end_ = (b); i <= i##_end_; ++ i)#define DREP(i, a, b) for(register int i = (a), i##_end_ = (b); i >= i##_end_; -- i)char buff[1<<25], *buf = buff;#define mem(a, b) memset((a), b, sizeof(a))template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }LL Fread(){    LL sum = 0, fg = 1;    for(; !isdigit(*buf) ; ++buf)if(*buf == '-')fg = -1;    for(; isdigit(*buf); ++buf)sum = sum * 10 + *buf - '0';    return sum * fg;}int read(){    int sum = 0,fg = 1;char c = getchar();    while(c < '0' || c > '9') { if (c == '-') fg = -1; c = getchar(); }    while(c >= '0' && c <= '9') { sum = sum * 10 + c - '0'; c = getchar(); }    return sum * fg;}const int maxn = 100000+10;const int inf = 0x3f3f3f3f;LL n,m;LL a[maxn],c1[maxn],c2[maxn];int lowbit(int x) {return (x & (-x));}void updata(int x,int num,int t){    if(t == 1)        for( ; x <= n; x += lowbit(x))            c1[x] += num;    else        for(; x <= n; x += lowbit(x))            c2[x] += num;}   LL query(int x,int t){    LL ans = 0;    if(t == 1)        for(; x > 0; x -= lowbit(x))            ans += c1[x];    else         for(; x > 0; x -= lowbit(x))            ans += c2[x];    return ans;}int main(){    fread(buff,1,1<<25,stdin);    n = Fread(), m = Fread();    REP(i,1,n) a[i] = Fread(), updata(i,a[i]-a[i-1],1),updata(i,(a[i]-a[i-1])*(i-1),2);    REP(i,1,m)    {        LL t = Fread();        if(t == 1)        {            LL x = Fread(), y = Fread(), k = Fread();            updata(x,k,1),updata(y + 1, -k, 1);            updata(x, (x-1) * k, 2), updata(y+1, -y * k, 2);        }        if(t == 2)        {            LL x = Fread(), y = Fread();            LL sum1 = query(x-1,1) * (x-1) - query(x-1,2);            LL sum2 = query(y,1) * y - query(y,2);            printf("%lld\n",sum2 - sum1);        }    }    return 0;}

逆序对

这里我们以权值为下标构建树状数组统计比当前点大的数的个数，边读边更新。

/*************************************************************************    > Author: Drinkwater-cnyali    > Created Time: 2017/5/25 10:51:38 ************************************************************************/#include<iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;#define REP(i, a, b) for(register int i = (a), i##_end_ = (b); i <= i##_end_; ++ i)#define DREP(i, a, b) for(register int i = (a), i##_end_ = (b); i >= i##_end_; -- i)#define debug(...) fprintf(stderr, __VA_ARGS__)#define mem(a, b) memset((a), b, sizeof(a))template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }int read(){    int sum = 0, fg = 1; char c = getchar();    while(c < '0' || c > '9') { if (c == '-') fg = -1; c = getchar(); }    while(c >= '0' && c <= '9') { sum = sum * 10 + c - '0'; c = getchar(); }    return sum * fg;}const int maxn = 100000;const int mod = 99999997;typedef pair<int,long long> P;P a[maxn],b[maxn];int n;int c[maxn],t[maxn];bool cmp(P u,P v){    return u.first < v.first;}int lowbit(int x){ return (x & (-x));}void updata(int x){    while(x <= n)        t[x]++,x += lowbit(x);}int get_sum(int x){    int ans = 0;    while(x > 0)    {        ans += t[x];        x -= lowbit(x);    }    return ans;}int main(){    n = read();    REP(i,1,n)a[i].first = read(),a[i].second = i;    REP(i,1,n)b[i].first = read(),b[i].second = i;    sort(a+1,a+1+n,cmp);sort(b+1,b+1+n,cmp);    REP(i,1,n)c[a[i].second] = b[i].second;    int ans = 0;    for(int i = n; i >= 1; --i)    {        ans = (ans%mod + get_sum(c[i])%mod)%mod;         updata(c[i]);    }    cout<<ans<<endl;    return 0;}