bzoj 3155: Preprefix sum 树状数组

题意

1<=N,M<=100000,且在任意时刻0<=Ai<=100000

分析

唉果然像我这种弱鸡就只能切切这种大家都会的sb题了。
把答案的式子展开后用树状数组维护一下对应项就好了。

代码

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int N=100005;int n,m,a[N];LL c1[N],c2[N];int read(){    int x=0,f=1;char ch=getchar();    while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}void ins1(int x,LL y){    while (x<=n) c1[x]+=y,x+=x&(-x);}void ins2(int x,LL y){    while (x<=n) c2[x]+=y,x+=x&(-x);}LL query1(int x){    LL ans=0;    while (x) ans+=c1[x],x-=x&(-x);    return ans;}LL query2(int x){    LL ans=0;    while (x) ans+=c2[x],x-=x&(-x);    return ans;}int main(){    n=read();m=read();    for (int i=1;i<=n;i++) a[i]=read(),ins1(i,a[i]),ins2(i,(LL)a[i]*i);    while (m--)    {        char ch[10];        scanf("%s",ch);        if (ch[0]=='M')        {            int x=read(),y=read();            ins1(x,-a[x]);ins2(x,(LL)-a[x]*x);            a[x]=y;            ins1(x,a[x]);ins2(x,(LL)a[x]*x);        }        else        {            int x=read();            LL s1=query1(x),s2=query2(x);            printf("%lld\n",(LL)s1*(x+1)-s2);        }    }    return 0;}