BZOJ Preprefix sum

Input
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
Output
对于每个询问操作，输出一行，表示所询问的SSi的值。
Sample Input
5 3
1 2 3 4 5
Query 5
Modify 3 2
Query 5
Sample Output
35
32
HINT
1<=N,M<=100000,且在任意时刻0<=Ai<=100000

设两个数组，一个数组res1维护ai， 一个数组res2维护(n-i+1)*ai;这里是由于S1 = a1；S2 = a1 + a2;S3 = a1 + a2 + a3;.....Sn = a1 + a2 + .... + an;则SSn =  ∑(n-i+1)*ai;求SSt =  getSum(res2, t)-(N-t)*getSum(res1, t)这里记得减去(N-t)*getSum(res1, t)举例：例如 求SS2： 我们可以清楚得算出SS2 = 2a1 + a2；而单独getSum(res2, t)算的却是SS2 = 5a1 + 4a2;这里比原先的值大了3a1 + 3a2， 是由于我们在维护的时候维护的是(n-i+1)*ai，长度是对应整个数组的长度，而这里是部分长，所以这里我们需要减去多余部分的(a1+a2),这里是多出的部分就是(5-2)*(a1+a2) = 3a1+3a2,所以我们写代码的时候是写成减去(n-i)*getSum(res1, i)；

#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cctype>#include<cmath>#include<ctime>#include<string>#include<stack>#include<deque>#include<queue>#include<list>#include<set>#include<map>#include<cstdio>#include<limits.h>#define MOD 1000000007#define fir first#define sec second#define fin freopen("/home/ostreambaba/文档/input.txt", "r", stdin)#define fout freopen("/home/ostreambaba/文档/output.txt", "w", stdout)#define mes(x, m) memset(x, m, sizeof(x))#define Pii pair<int, int>#define Pll pair<ll, ll>#define INF 1e9+7#define Pi 4.0*atan(1.0)#define lowbit(x) (x&(-x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define scini(n) scanf("%d", &n)#define scinl(n) scanf("%lld", &n)#define scinii(n, m) scanf("%d%d", &n, &m)#define scinll(n, m) scanf("%lld%lld", &n, &m)#define scout(n) printf("%d\n", n);typedef long long ll;typedef unsigned long long ull;const double eps = 1e-6;const int maxn = 50010;using namespace std;int N, M;ll in[100100];ll res1[100100];ll res2[100100];ll a, b;void Modify(ll *S, int i, ll x){    while(i <= N){        S[i] += x;        i += lowbit(i);    }}ll getSum(ll *S, int i){    ll ret = 0;    while(i > 0){        ret += S[i];        i -= lowbit(i);    }    return ret;}void Query(int k){    printf("%lld\n", getSum(res2, k)-(N-k)*getSum(res1, k));}int main(){    //fin;    scinii(N, M);    mes(res1, 0);    mes(res2, 0);    for(int i = 1; i <= N; ++i){        scinl(in[i]);        Modify(res1, i, in[i]);        Modify(res2, i, (N-i+1)*in[i]);    }    char op[7];    while(M--){        scanf("%s", op);        if(op[0] == 'M'){            scinll(a, b);            Modify(res1, a, b - in[a]);  //要先减取之前加的数，加上新的数            Modify(res2, a, (N-a+1)*(b - in[a]));            in[a] = b;        }        else{            scinl(a);            Query(a);        }    }    return 0;}