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本题链接（AcWing） 点这里

题目描述

给定一个长度为 $N$ 的数列 $A$，以及 $M$ 条指令，每条指令可能是以下两种之一：

1. C l r d，表示把 $A[l],A[l+1],\dots ,A[r]$ 都加上 $d$。
2. Q l r，表示询问数列中第 $l\sim r$ 个数的和。

对于每个询问，输出一个整数表示答案。

输入格式

第一行两个整数 $N,M$。

第二行 $N$ 个整数 $A[i]$。

接下来 $M$ 行表示 $M$ 条指令，每条指令的格式如题目描述所示。

输出格式

对于每个询问，输出一个整数表示答案。

每个答案占一行。

数据范围

$1\le N,M\le 10^5$,
$|d|\le 10000$,
$|A[i]|\le 10^9$

输入样例：

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

输出样例：

4
55
9
15

思路

考虑用树状数组 tr[] 维护差分数组

则求原数组的前缀和

$$\{\begin{array}{cccccccccc}{a}_1& =& d_1& & & & & & & \\ a_2& =& d_1& +& d_2& & & & & \\ a_3& =& d_1& +& d_2& +& d_3& & & \\ .& .& .& .& .& .& & & & \\ a_n& =& d_1& +& d_2& +& ...& +& d_n& \end{array}$$

$$s_i=\sum _{i=1}^n{a}_i=\{\begin{array}{cccccccccc}{d}_1& & & & & & & & & \\ d_1& +& d_2& & & & & & & \\ d_1& +& d_2& +& d_3& & & & & \\ .& .& .& .& .& .& & & & \\ d_1& +& d_2& +& ...& +& d_n& & & \end{array}$$

我们考虑把后面的矩阵补全：

则

$$s_i=(n+1)\times \sum _{i=1}^n{d}_i-\sum _{i=1}^n(i\times d_i)$$

所以我们需要两个树状数组，tr1[] 维护差分数组，tr2[] 维护 $i\times d_i$

代码：

#include <iostream>using namespace std;typedef long long LL;const LL N = 1000010;LL n, m;
LL a[N];
LL t1[N], t2[N];inline LL lowbit(LL x)
{return x & -x;
}inline void add(LL t[], LL x, LL c)
{for (LL i = x; i <= n; i += lowbit(i))t[i] += c;
}inline LL sum(LL t[], LL x)
{LL res = 0;for (LL i = x; i; i -= lowbit(i))res += t[i];return res;
}inline LL psum(LL x)
{return sum(t1, x) * (x + 1) - sum(t2, x);
}int main()
{scanf("%lld%lld", &n, &m);for (LL i = 1; i <= n; i ++ ) scanf("%lld", &a[i]);for (LL i = 1; i <= n; i ++ ){LL b = a[i] - a[i - 1];add(t1, i, b);add(t2, i, b * i);}while (m -- ){char op[2];LL l, r, d;scanf("%s%lld%lld", op, &l, &r);if (op[0] == '2'){printf("%lld\n", psum(r) - psum(l - 1));}else{scanf("%lld", &d);add(t1, l, d), add(t2, l, l * d);add(t1, r + 1, -d), add(t2, r + 1, -d * (r + 1));}}return 0;
}

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（后附极限卡常代码，70ms，较优解）

#define qwq optimize
#pragma GCC qwq(1)
#pragma GCC qwq(2)
#pragma GCC qwq(3)
#pragma GCC qwq("Ofast")
#pragma GCC qwq("inline")
#pragma GCC qwq("-fgcse")
#pragma GCC qwq("-fgcse-lm")
#pragma GCC qwq("-fipa-sra")
#pragma GCC qwq("-ftree-pre")
#pragma GCC qwq("-ftree-vrp")
#pragma GCC qwq("-fpeephole2")
#pragma GCC qwq("-ffast-math")
#pragma GCC qwq("-fsched-spec")
#pragma GCC qwq("unroll-loops")
#pragma GCC qwq("-falign-jumps")
#pragma GCC qwq("-falign-loops")
#pragma GCC qwq("-falign-labels")
#pragma GCC qwq("-fdevirtualize")
#pragma GCC qwq("-fcaller-saves")
#pragma GCC qwq("-fcrossjumping")
#pragma GCC qwq("-fthread-jumps")
#pragma GCC qwq("-funroll-loops")
#pragma GCC qwq("-fwhole-program")
#pragma GCC qwq("-freorder-blocks")
#pragma GCC qwq("-fschedule-insns")
#pragma GCC qwq("inline-functions")
#pragma GCC qwq("-ftree-tail-merge")
#pragma GCC qwq("-fschedule-insns2")
#pragma GCC qwq("-fstrict-aliasing")
#pragma GCC qwq("-fstrict-overflow")
#pragma GCC qwq("-falign-functions")
#pragma GCC qwq("-fcse-skip-blocks")
#pragma GCC qwq("-fcse-follow-jumps")
#pragma GCC qwq("-fsched-interblock")
#pragma GCC qwq("-fpartial-inlining")
#pragma GCC qwq("no-stack-protector")
#pragma GCC qwq("-freorder-functions")
#pragma GCC qwq("-findirect-inlining")
#pragma GCC qwq("-fhoist-adjacent-loads")
#pragma GCC qwq("-frerun-cse-after-loop")
#pragma GCC qwq("inline-small-functions")
#pragma GCC qwq("-finline-small-functions")
#pragma GCC qwq("-ftree-switch-conversion")
#pragma GCC qwq("-fqwq-sibling-calls")
#pragma GCC qwq("-fexpensive-optimizations")
#pragma GCC qwq("-funsafe-loop-optimizations")
#pragma GCC qwq("inline-functions-called-once")
#pragma GCC qwq("-fdelete-null-pointer-checks")
#include <iostream>
#include <cstdio>#define lb(x) (x & (-x))using namespace std;typedef long long LL;const LL N = 100010;LL n, m;
LL a[N];
LL t1[N], t2[N];char *p1, *p2, buf[N];
#define nc() (p1 == p2 && (p2 = (p1 = buf) +\
fread(buf, 1, N, stdin), p1 == p2) ? EOF : *p1 ++ )
LL read()
{LL x = 0, f = 1;char ch = nc();while (ch < 48 || ch > 57){if (ch == '-') f = -1;ch = nc();}while (ch >= 48 && ch <= 57)x = (x << 3) + (x << 1) + (ch ^ 48), ch = nc();return x * f;
}char obuf[N], *p3 = obuf;
#define putchar(x) (p3 - obuf < N) ? (*p3 ++ = x) :\
(fwrite(obuf, p3 - obuf, 1, stdout), p3 = obuf, *p3 ++ = x)
inline void write(LL x)
{if (!x){putchar('0');return;}LL len = 0, k1 = x, c[40];if (k1 < 0) k1 = -k1, putchar('-');while (k1) c[len ++ ] = k1 % 10 ^ 48, k1 /= 10;while (len -- ) putchar(c[len]);
}inline void add(LL t[], LL x, LL c)
{for (LL i = x; i <= n; i += lb(i))t[i] += c;
}inline LL sum(LL t[], LL x)
{LL res = 0;for (LL i = x; i; i -= lb(i))res += t[i];return res;
}inline LL psum(LL x)
{return sum(t1, x) * (x + 1) - sum(t2, x);
}int main()
{n = read(), m = read();for (LL i = 1; i <= n; i ++ ) a[i] = read();for (LL i = 1; i <= n; i ++ ){LL b = a[i] - a[i - 1];add(t1, i, b);add(t2, i, b * i);}LL op, l, r, d;while (m -- ){op = read(), l = read(), r = read();if (op == 2) write(psum(r) - psum(l - 1)), putchar(10);else{d = read();add(t1, l, d), add(t2, l, l * d);add(t1, r + 1, -d), add(t2, r + 1, -d * (r + 1));}}fwrite(obuf, p3 - obuf, 1, stdout);return 0;
}

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