【bzoj3155】Preprefix Sum

　　这题目名字挺有意思= =
　　 SSt=∑ti=1ai∗i=∑ti=1ai∗(n−i+1)−(n−t)∑ti=1ai
　　 维护两个前缀和， ∑(n−i+1)ai 和 ∑ai ，然后直接查询、修改即可。
　　 时间复杂度 O(nlogn)

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
#define lowbit(x) (x & (-x))
#define maxn 100007

inline int rd() {
    char c = getchar();
    while (!isdigit(c)) c = getchar() ; int x = c - '0';
    while (isdigit(c = getchar())) x = x * 10 + c - '0';
    return x;
}

typedef long long ll;

int n , m;
ll  a[maxn];

struct BITS {
    ll val[maxn];

    void add(int x , ll v) {
        while (x <= n)
            val[x] += v , x += lowbit(x);
    }

    ll sum(int x) {
        ll t = 0;
        while (x)
            t += val[x] , x -= lowbit(x);
        return t;
    }
}s1 , s2;

void input() {
    n = rd() , m = rd();
    rep (i , 1 , n) {
        a[i] = rd();
        s1.add(i , a[i]);
        s2.add(i , (ll) (n - i + 1) * a[i]);
    }
}

void solve() {
    char cmd[10];
    rep (i , 1 , m) {
        scanf("%s" , cmd);
        int t = rd();
        if (cmd[0] == 'Q') {
            printf("%lld\n" , s2.sum(t) - (n - t) * s1.sum(t));
        }
        else {
            int v = rd() - a[t];
            s1.add(t , v) , s2.add(t , (ll) (n - t + 1) * v);
            a[t] += v;
        }
    }
}

int main() {
    #ifndef ONLINE_JUDGE
        freopen("data.txt" , "r" , stdin);
    #endif
    input();
    solve();
    return 0;
}