POJ-3468-A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 93547 Accepted: 29145Case Time Limit: 2000MS

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 a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" 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.

线段树, 板子题, 注意区间更新是在原基础上加

#include<stdio.h>#include<string.h>#include<math.h>//线段树//用途：快速区间更新，区间查询最值或和//初始化（多组数据）：Initialize()//注意事项：所有函数调用时前3个参数均填写为：1, 1, n，其他参数下方具体描述。默认为单点修改，区间求和，其他需求请自行更改int n, q;const int MAX = 200000+1;//区间更新、区间查询（只能查询区间和）//il、ir分别为目标区间的左、右端点下标，value为要更新的值long long  st[MAX*6];long long lazy[MAX*6];void Initialize(){    memset(st, 0, sizeof(st));    memset(lazy, 0, sizeof(lazy));}void Update(int node, int l, int r, int il, int ir, long long value){    int mid = (l + r) >> 1;    if(lazy[node])    {        lazy[node<<1] += lazy[node];        lazy[node<<1|1] += lazy[node];        st[node<<1] += (mid-l+1) * lazy[node];        st[node<<1|1] += (r-mid) * lazy[node];        lazy[node] = 0;    }    if(l==il && r==ir)    {        lazy[node] += value;        st[node] += (r-l+1) * value;        return;    }    if(ir <= mid)        Update(node<<1, l, mid, il, ir, value);    else if(il > mid)        Update(node<<1|1, mid+1, r, il, ir, value);    else    {        Update(node<<1, l, mid, il, mid, value);        Update(node<<1|1, mid+1, r, mid+1, ir, value);    }    st[node] = st[node<<1] + st[node<<1|1];}long long Query(int node, int l, int r, int il, int ir){    int mid = (l + r) >> 1;    if(lazy[node])    {        lazy[node<<1] += lazy[node];        lazy[node<<1|1] += lazy[node];        st[node<<1] += (mid-l+1) * lazy[node];        st[node<<1|1] += (r-mid) * lazy[node];        lazy[node] = 0;    }    if(l==il && r==ir)        return st[node];    if(ir <= mid)        return Query(node<<1, l, mid, il, ir);    else if(il > mid)        return Query(node<<1|1, mid+1, r, il, ir);    else        return Query(node<<1, l, mid, il, mid) +               Query(node<<1|1, mid+1, r, mid+1, ir);}int main(){    while(~scanf("%d%d", &n,&q))    {        Initialize();        int  a, b;        long long x;        for(int i = 1; i<=n; ++i)        {            scanf("%lld", &x);            Update(1, 1, n, i, i,  x);        }        char s[2];        while(q--)        {            scanf("%s", s);            if(s[0]=='Q')            {                scanf("%d%d",&a, &b);                if(a>b)                {                    a+=b;                    b=a-b;                    a-=b;                }                printf("%lld\n", Query(1,1,n,a,b));            }            else if(s[0]=='C')            {                scanf("%d%d%lld", &a, &b,&x);                if(a>b)                {                    a+=b;                    b=a-b;                    a-=b;                }                Update(1, 1, n, a, b, x);            }        }    }    return 0;}