poj3468 A Simple Problem with Integers

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#include <iostream>#include <cstdio>#include <cstdlib>//基於線段樹的區間維護//維護結點o，它相應區間[L,R]#define ll long longconst int MAX_N=111111;const int MAX_Q=111111;int N,Q;int A[MAX_N+1];char T[MAX_Q];int L[MAX_Q],R[MAX_Q],X[MAX_Q];ll bit0[MAX_N+1],bit1[MAX_N+1];ll sum(ll *b,int i){    ll s=0;    while(i>0){        s+=b[i];        i-=i & -i;    }    return s;}void add(ll* b,int i,int v){    while(i<=N)    {        b[i]+=v;        i+=i&-i;    }}int main(){    #ifndef ONLINE_JUDGEfreopen("D:/1.txt","r",stdin);freopen("D:/2.txt","w",stdout);#endif    scanf("%d%d",&N,&Q);    for(int i=1;i<=N;i++)    {        scanf("%d",&A[i]);    }    for(int i=1;i<=Q;i++)    {        char ch[2];        scanf("%s",ch);        if(ch[0]=='Q')        {            T[i]=ch[0];            scanf("%d%d",&L[i],&R[i]);        }        else        {            T[i]=ch[0];            scanf("%d%d%d",&L[i],&R[i],&X[i]);        }    }    for(int i=1;i<=N;i++)    {        add(bit0,i,A[i]);    }    for(int i=1;i<=Q;i++)    {        if(T[i]=='C')        {            add(bit0,L[i],-X[i]*(L[i]-1));            add(bit1,L[i],X[i]);            add(bit0,R[i]+1,X[i]*R[i]);            add(bit1,R[i]+1,-X[i]);        }        else        {            ll res=0;            res+=sum(bit0,R[i])+sum(bit1,R[i])*R[i];            res-=sum(bit0,L[i]-1)+sum(bit1,L[i]-1)*(L[i]-1);            printf("%lld\n",res);        }    }}

 

poj3468 A Simple Problem with Integers