【POJ 3468】 A Simple Problem with Integers

【题目链接】

点击打开链接

【算法】

本题用线段树很容易写，但是，笔者为了练习树状数组，就用树状数组的方法做了一遍

我们不妨引入差分数组c，

则sum(n) = c[1] + (c[1] + c[2]) + (c[1] + c[2] + c[3]) + ... + (c[1] + c[2] + c[3] + ... + c[n])

= n * c[1] + (n - 1) * c[2] + (n - 2) * c[3] + ... + c[n]

= n * (c[1] + c[2] + c[3] + ... + c[n]) - c[2] - c[3] * 2 - c[4] * 3 - ... - c[n] * (n - 1)

所以可以用两个树状数组分别维护c的前缀和和c[i]*(i-1)的前缀和

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 100000

long long N,Q,tr,tl,i,l,r,x;
long long a[MAXN+];
char opt;

template <typename T> inline void read(T &x) {
    long long f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
template <typename T> inline void write(T x) {
    if (x < ) { x = -x; putchar('-'); }
    if (x > ) write(x/);
    putchar(x%+'');
}
template <typename T> inline void writeln(T x) {
    write(x);
    puts("");
} 

struct BinaryIndexedTree {
        long long bit[MAXN+];
        inline long long lowbit(long long x) { return x & -x; }
        inline void clear() {
                long long i;
                for (i = ; i <= N; i++) bit[i] = ;
        }
        inline void modify(long long pos,long long val) {
                long long i;
                for (i = pos; i <= N; i += lowbit(i)) bit[i] += val;
        }
        inline long long query(long long pos) {
                long long i,ret = ;
                for (i = pos; i; i -= lowbit(i)) ret += bit[i];
                return ret;
        }
} c1,c2;

int main() {

        read(N); read(Q);
        for (i = ; i <= N; i++) {
                read(a[i]);
                c1.modify(i,a[i]);
                c1.modify(i+,-a[i]);
                c2.modify(i,(i-)*a[i]);
                c2.modify(i+,-i*a[i]);
        }
        while (Q--) {
                opt = getchar();
                if (opt == 'C') {
                        read(l); read(r); read(x);
                        c1.modify(l,x);
                        c1.modify(r+,-x);
                        c2.modify(l,(l-)*x);
                        c2.modify(r+,-x*r);
                } else {
                        read(l); read(r);
                        tl = c1.query(l-) * (l - ) - c2.query(l-);
                        tr = c1.query(r) * r - c2.query(r);
                        writeln(tr-tl);
                }
        }

        return ;
}
/*
   c[1] + (c[1] + c[2]) + (c[1] + c[2] + c[3]) + ... + (c[1] + c[2] + c[3] + c[4] + ... + c[n])
   = c[1] * n + c[2] * (n - 1) + c[3] * (n - 2) + ... + c[n]
   = (c[1] + c[2] + c[3] + ... + c[n]) * n - c[2] - c[3] * 2 - c[4] * 3 - ... - c[n] * (n - 1)
   = sigma(c1,n) * n - sigma(c2,n)
*/