2018年全国多校算法寒假训练营练习比赛（第五场）题解 H Tree Recovery

Tree Recovery

链接：https://www.nowcoder.com/acm/contest/77/H
来源：牛客网

Tree Recovery时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 131072K，其他语言262144K
64bit IO Format: %lld

题目描述

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.

输入描述:

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.

输出描述:

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

示例1

输入

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

输出

4
55
9
15

AC代码：# include<stdio.h>
int main()
{
long long int a[100000],e;
long long int i,j,n,m,x,y,h;
char t;
scanf("%lld%lld",&n,&m);
for(i=0;i<n;i++)
scanf("%lld",&a[i]);
for(i=0;i<m;i++)
{
scanf("%c",&t);
scanf("%c",&t);
if(t=='Q')
{
scanf("%lld%lld",&x,&y);
e=0;
for(j=x-1;j<y;j++)
e=e+a[j];
printf("%lld\n",e);
}
else
{
scanf("%lld%lld%lld",&x,&y,&h);
for(j=x-1;j<y;j++)
a[j]=a[j]+h;
}
}
return 0;
}