POJ3468 线段树模板

Description

You have N integers, A₁, A₂, ... ,A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁,A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C abc" means adding c to each of A_a, A_a+1, ... ,A_b. -10000 ≤ c ≤ 10000.
"Q ab" means querying the sum of A_a, A_a+1, ... ,A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

The sums may exceed the range of 32-bit integers.


这个题之前做过，模板题
郁闷的是今天找了一晚上，自己写的也没找出WA在哪。。。。
今天的WA

#include <iostream>#include<stdio.h>using namespace std;const int N=1e5;typedef long long ll;struct node{    int left,right;    ll weight;    int flag;}tree[4*N+500];ll sum;void build(int k,int ll,int rr){    tree[k].left=ll;    tree[k].right=rr;    tree[k].flag=0;    if(tree[k].left==tree[k].right)    {        scanf("%lld",&tree[k].weight);        tree[k].flag=0;        return ;    }    int mid=(ll+rr)/2;    build(2*k,ll,mid);    build(2*k+1,mid+1,rr);    tree[k].weight=tree[2*k].weight+tree[2*k+1].weight;}void down(int k){    tree[2*k].flag+=tree[k].flag;    tree[2*k+1].flag+=tree[k].flag;    tree[2*k].weight+=tree[k].flag*(tree[2*k].right-tree[2*k].left+1);    tree[2*k+1].weight+=tree[k].flag*(tree[2*k+1].right-tree[2*k+1].left+1);    tree[k].flag=0;}void ask_interval(int k,int a,int b){    if(tree[k].left>=a&&tree[k].right<=b)    {        sum+=tree[k].weight;        return ;    }    if(tree[k].flag)        down(k);    int mid=(tree[k].left+tree[k].right)/2;    if(a<=mid)        ask_interval(2*k,a,b);    if(b>mid)        ask_interval(2*k+1,a,b);}void change_interval(int k,int a,int b,int add){    if(tree[k].left>=a&&tree[k].right<=b)    {        tree[k].weight+=(tree[k].right-tree[k].left+1)*add;        tree[k].flag+=add;        return ;    }    if(tree[k].flag)        down(k);    int mid=(tree[k].left+tree[k].right)/2;    if(a<=mid)        change_interval(2*k,a,b,add);    if(b>mid)        change_interval(2*k+1,a,b,add);    tree[k].weight=tree[2*k].weight+tree[2*k+1].weight;}int main(){   int m,n;    while(scanf("%d%d",&n,&m)==2&&n+m!=0)   {       build(1,1,n);        int u,v;       char c;       int add;       while(m--)       {           getchar();           c=getchar();           scanf("%d%d",&u,&v);           if(c=='Q')           {               scanf("%d%d",&u,&v);               sum=0;               ask_interval(1,u,v);               printf("%lld\n",sum);           }           if(c=='C')           {                scanf("%d%d%d",&u,&v,&add);               change_interval(1,u,v,add);           }       }   }    return 0;}

之前的AC

#include <iostream>#include<stdio.h>using namespace std;typedef long long ll;const int N=1e5+50;struct node{    ll left,right;    ll weight;    int flag;}tree[4*N];ll fn[N];int o,n,q;ll ans;void build(ll k,ll l,ll r){    tree[k].left=l;    tree[k].right=r;    if(tree[k].left==tree[k].right){        tree[k].weight=fn[o++];        tree[k].flag=0;        return ;    }    ll mid=(l+r)/2;    build(2*k,l,mid);    build(2*k+1,mid+1,r);    tree[k].weight=tree[2*k].weight+tree[2*k+1].weight;}void down(ll k){    tree[2*k].flag+=tree[k].flag;    tree[2*k+1].flag+=tree[k].flag;    tree[2*k].weight+=tree[k].flag*(tree[2*k].right-tree[2*k].left+1);    tree[2*k+1].weight+=tree[k].flag*(tree[2*k+1].right-tree[2*k+1].left+1);    tree[k].flag=0;}void ask_interval(ll k,ll a,ll b){    if(tree[k].left>=a&&tree[k].right<=b){        ans+=tree[k].weight;        return ;    }    if(tree[k].flag)        down(k);    ll mid=(tree[k].left+tree[k].right)/2;    if(a<=mid)        ask_interval(2*k,a,b);    if(b>mid)        ask_interval(2*k+1,a,b);}void change_interval(ll k,ll a,ll b,int add){    if(tree[k].left>=a&&tree[k].right<=b){        tree[k].weight+=(tree[k].right-tree[k].left+1)*add;        tree[k].flag+=add;        return ;    }    if(tree[k].flag)        down(k);    ll mid=(tree[k].left+tree[k].right)/2;    if(a<=mid)        change_interval(2*k,a,b,add);    if(b>mid)        change_interval(2*k+1,a,b,add);    tree[k].weight=tree[k*2].weight+tree[k*2+1].weight;}int main(){    while(scanf("%d%d",&n,&q)==2&&n+q!=0)   {        o=1;        for(int i=1;i<=n;i++)            scanf("%lld",&fn[i]);        build(1,1,n);        ll x,y;        int z;        char c;        while(q--){            getchar();            c=getchar();            if(c=='C'){                scanf("%lld%lld%d",&x,&y,&z);                change_interval(1,x,y,z);            }            if(c=='Q'){                ans=0;                scanf("%lld%lld",&x,&y);                ask_interval(1,x,y);                printf("%lld\n",ans);            }        }   }    return 0;}

今天先到这吧，不开心~