1004:Challenge 4

总时间限制: 

10000ms

单个测试点时间限制: 

1000ms

内存限制: 

262144kB

描述

给一个长为N的数列，有M次操作，每次操作时以下三种之一：

（1）修改数列中的一个数

（2）求数列中某连续一段所有数的两两乘积的和 mod 1000000007

（3）求数列中某连续一段所有相邻两数乘积的和 mod 1000000007

输入

第一行两个正整数N和M。
第二行N的整数表示这个数列。
接下来M行，每行开头是一个字符，若该字符为'M'，则表示一个修改操作，接下来两个整数x和y，表示把x位置的值修改为y；若该字符为'Q'，则表示一个询问操作，接下来两个整数x和y，表示对[x,y]区间做2号询问；若该字符为'A'，则表示一个询问操作，接下来两个整数x和y，表示对[x,y]区间做3号询问。

输出

对每一个询问操作单独输出一行，表示答案。

样例输入

5 5
1 2 3 4 5
Q 1 5
A 1 5
M 2 7
Q 1 5
A 1 5

样例输出

85
40
150
60

提示

1<=N<=10^5，1<=M<=10^5，输入保证合法，且所有整数可用带符号32位整型存储。

#include<iostream>
#include<cstdio>
#define ll long long
using namespace std;const int mod = 1000000007;
const int N = 100005;ll a[N];
int n, m;struct node {node *ls, *rs;ll sum, num1, num2, lf, rg;
} pool[N*32], *tail = pool, *zero, *root;///指针池大小void update ( node *nd, node *ld, node *rd ) {nd -> sum = ( ld -> sum + rd -> sum + mod ) % mod;nd -> num1 = (  ld -> sum * rd -> sum  % mod + ld -> num1 + rd -> num1 + mod ) % mod;nd -> num2 = ( ld -> num2 + rd -> num2 + ld -> rg * rd -> lf % mod + mod ) % mod;nd -> lf = ld -> lf % mod; nd -> rg = rd -> rg % mod;
}node *build ( int l, int r ) {node *nd = ++ tail;if ( l == r ) {ll pp = ( a[l] % mod + mod ) % mod;nd -> lf = nd -> rg = pp;nd -> sum = pp; nd -> num1 = nd -> num2 = 0;return nd;}int mid = ( l + r ) >> 1;nd -> ls = build ( l, mid );nd -> rs = build ( mid + 1, r );update ( nd, nd -> ls, nd -> rs );return nd;
}node *query ( node *nd, int l, int r, int L, int R ) {if ( l >= L && r <= R )return nd;int mid = ( l + r ) >> 1;node *nd1 = ++ tail, *nd2 = ++ tail;if ( mid >= L )nd1 = query ( nd -> ls, l, mid, L, R );if ( mid < R )nd2 = query ( nd -> rs, mid + 1, r, L, R );node *st = ++ tail;update ( st, nd1, nd2 );return st;
}void modify ( node *nd, int l, int r, int pos, ll d ) {if ( l == r ) {nd -> lf = nd -> rg = d;nd -> sum = d;return ;}int mid = ( l + r ) >> 1;if ( pos <= mid ) modify ( nd -> ls, l, mid, pos, d );else modify ( nd -> rs, mid + 1, r, pos, d );update ( nd, nd -> ls, nd -> rs );
}int main ( ) {scanf ( "%d%d", &n, &m );for ( int i = 1; i <= n; i ++ )scanf ( "%lld", &a[i] );root = build ( 1, n );for ( int i = 1; i <= m; i ++ ) {char opt;scanf ( "\n%c", &opt );if ( opt == 'Q' ) {int l, r;scanf ( "%d%d", &l, &r );node *ans = query ( root, 1, n, l, r );printf ( "%lld\n", ( ans -> num1 + mod ) % mod );} else if ( opt == 'A' ) {int l, r;scanf ( "%d%d", &l, &r );node *ans = query ( root, 1, n, l, r );printf ( "%lld\n", ( ans -> num2 + mod ) % mod );} else {int pos; ll r;scanf ( "%d%lld", &pos, &r );r = ( r + mod ) % mod;////modify ( root, 1, n, pos, r );}}return 0;
}