题目描述
前缀和(prefix sum)Si=∑k=1iaiS_i=\sum_{k=1}^i a_iSi=∑k=1iai。
前前缀和(preprefix sum) 则把SiS_iSi作为原序列再进行前缀和。记再次求得前缀和第i个是SSiSS_iSSi
给一个长度n的序列a1,a2,⋯,ana_1, a_2, \cdots, a_na1,a2,⋯,an，有两种操作：
-
Modify i x：把aia_iai改成xxx； -
Query i：查询SSiSS_iSSi
输入输出格式
输入格式：
第一行给出两个整数N，M。分别表示序列长度和操作个数
接下来一行有N个数，即给定的序列a1,a2,....an
接下来M行，每行对应一个操作，格式见题目描述
输出格式：
对于每个询问操作，输出一行，表示所询问的SSi的值。
输入输出样例
说明
1<=N,M<=100000,且在任意时刻0<=Ai<=100000
维护两个数组：a[i], (n-i+1)a[i];
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-11
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii; inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
} ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; } /*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/ ll a[maxn];
ll b[2][maxn]; int n, m; void add(int x, ll val, int dx) {
while (x <= n) {
b[dx][x] += val; x += x & -x;
}
}
void upd(int pos, ll val) {
add(pos, -a[pos], 0); add(pos, -a[pos] * (n - pos + 1), 1);
a[pos] = val;
add(pos, a[pos], 0); add(pos, a[pos] * (n - pos + 1), 1);
}
ll query(int x) {
ll ans = 0;
int tmp = x;
while (tmp > 0) {
ans += b[1][tmp]; tmp -= tmp & -tmp;
}
tmp = x; ll res = 0;
while (tmp > 0) {
res += b[0][tmp]; tmp -= tmp & -tmp;
}
return ans - (n - x)*res;
} int main() {
// ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
n = rd(); m = rd();
for (int i = 1; i <= n; i++) {
rdllt(a[i]);
add(i, a[i], 0); add(i, (n - i + 1)*a[i], 1);
}
while (m--) {
char ch[10];
rdstr(ch);
if (ch[0] == 'M') {
int pos; ll val;
rdint(pos); rdllt(val);
upd(pos, val);
}
else {
int pos; pos = rd();
printf("%lld\n", query(pos));
}
}
return 0;
}