Foreword
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answer
We observed the expression
\ [AVG = \ n-FRAC {} {}. 1 \ sum_ = {I}. 1 ^ n-a_i \]
\ [D = \ {n-FRAC {}}. 1 \ sum_ = {I}. 1 ^ n- (a_i-avg) ^ 2 \
] we expand it into
\ [d = \ frac {1 } {n} \ sum_ {i = 1} ^ n (a_i ^ 2-2 \ times a_i \ times avg - avg ^ 2) \]
proposed to obtain a constant term
\ [d = \ frac {1 } {n} (\ sum_ {i = 1} ^ n a_i ^ 2-2 \ times avg \ times \ sum_ {i = 1} ^ n a_i + n \ times avg ^ 2
) \] then we opened two tree line, a maintenance \ (a_i \) and, a maintenance \ (a_i ^ 2 \) and, in addition a multiplicative inverse , ie can.
Bloggers also result konjac WA once, becauseModulo sedentary
Code
Two tree line on the deal together, is very convenient.
Inverse statement in accordance with the giant man, with a quick power (Goodbye Top Euro).
#include <cstdio>
#define ll long long
ll s[400005], s2[400005];
const ll MOD = 1e9 + 7;
ll read(){
ll x = 0; int zf = 1; char ch = ' ';
while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;
}
ll getInv(ll x){
ll res = 1;
for (int y = MOD - 2; y; y >>= 1, x = (x * x) % MOD)
if (y & 1)
(res *= x) %= MOD;
return res;
}
void build(int pos, int l, int r){
if (l == r){
s[pos] = read(); s2[pos] = (s[pos] * s[pos]) % MOD;
return ;
}
int mid = (l + r) >> 1;
build(pos << 1, l, mid);
build(pos << 1 | 1, mid + 1, r);
s[pos] = s[pos << 1] + s[pos << 1 | 1];
s2[pos] = s2[pos << 1] + s2[pos << 1 | 1];
}
ll query(int pos, int l, int r, int x, int y){
if (x <= l && r <= y)
return s[pos];
ll ans = 0; int mid = (l + r) >> 1;
if (x <= mid)
ans += query(pos << 1, l, mid, x, y);
if (mid < y)
(ans += query(pos << 1 | 1, mid + 1, r, x, y)) %= MOD;
return ans;
}
ll query2(int pos, int l, int r, int x, int y){
if (x <= l && r <= y)
return s2[pos];
ll ans = 0; int mid = (l + r) >> 1;
if (x <= mid)
ans += query2(pos << 1, l, mid, x, y);
if (mid < y)
(ans += query2(pos << 1 | 1, mid + 1, r, x, y)) %= MOD;
return ans;
}
void modify(int pos, int l, int r, int x, ll val){
if (l == r){
s[pos] = val, s2[pos] = (val * val) % MOD;
return ;
}
int mid = (l + r) >> 1;
if (x <= mid)
modify(pos << 1, l, mid, x, val);
else if (mid < x)
modify(pos << 1 | 1, mid + 1, r, x, val);
s[pos] = (s[pos << 1] + s[pos << 1 | 1]) % MOD;
s2[pos] = (s2[pos << 1] + s2[pos << 1 | 1]) % MOD;
}
int main(){
int n = read(), m = read();
build(1, 1, n); ll c, a, b;
while (m--){
c = read(), a = read(), b = read();
if (c == 1)
modify(1, 1, n, a, b);
else if (c == 2){
ll ans2 = query2(1, 1, n, a, b), ans = query(1, 1, n, a, b);
ll avg = (ans * getInv(b - a + 1)) % MOD;
ll res = (((ans2 - (avg * ans * 2ll % MOD) + ((((avg * avg) % MOD) * (b - a + 1)) % MOD)) % MOD * getInv(b - a + 1))) % MOD;
while (res < 0) res += MOD;
printf("%lld\n", res);
}
}
return 0;
}