【POJ 3468 A Simple Problem with Integers】 线段树

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 115961 Accepted: 36028
Case Time Limit: 2000MS
Description

You have N integers, A1, A2, … , AN. 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, … , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
“C a b c” means adding c to each of Aa, Aa+1, … , Ab. -10000 ≤ c ≤ 10000.
“Q a b” means querying the sum of Aa, Aa+1, … , Ab.

Output

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

Sample Input

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

4
55
9
15
Hint

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

线段树 & 比较巧妙的是用了另一个数组记录节点区间的更新，查询从而更高效了

AC代码：

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int MAX = 4e5 + 10;typedef long long LL;LL n,m,a[MAX],sum[MAX],x[MAX];LL init(int l,int r,int k){    if(l == r) return sum[k] = a[l];    int o = l + r >> 1;    return sum[k] += init( l,o,k * 2) + init(o + 1,r,k * 2 + 1);}void ch(int L,int R,int l,int r,int k,int w){    if(l > R || r < L) return ;    else if(l <= L && R <= r){        x[k] += w; return ;    }    int o = L + R >> 1;    ch(L,o,l,r,k * 2,w),ch(o + 1,R,l,r,k * 2 + 1,w);    sum[k] += (LL)(min(R,r) - max(L,l) + 1) * w;}LL qu(int L,int R,int l,int r,int k){    if(l > R || r < L) return 0;    else if(l <= L && R <= r) return sum[k] + (LL)x[k] * (R - L + 1);    int o = L + R >> 1;    return qu(L,o,l,r,k * 2) + qu(o + 1,R,l,r,k * 2 + 1) + (LL)(min(R,r) - max(L,l) + 1) * x[k];}int main(){    scanf("%lld %lld",&n,&m);    for(int i = 1; i <= n; i++)        scanf("%lld",&a[i]);    init(1,n,1);    while(m--){        char s[2];        int l,r,o; scanf("%s %d %d",s,&l,&r);        if(s[0] == 'C') scanf("%d",&o),ch(1,n,l,r,1,o);        else printf("%lld\n",qu(1,n,l,r,1));    }    return 0;}