A Simple Problem with Integers （线段树）

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.

给定一个长为n的序列，接下来m次操作，每次操作会对序列某段区间内所有数加上一个值，或是询问一段区间的和。n，m≤10^5。

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线段树的模板，刚开始死活A不了，后来发现输入的判定有问题

这道题的输入不是0，1，2

而是一串字符，取最开始的一个，但是后面的也要输入

坑啊！！！

下面给出代码：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
inline long long rd(){
    long long x=0,f=1;
    char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-1;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
    return x*f;
}
inline void write(long long x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
    return ;
}
long long n,m;
struct node{
    long long sum,l,r;
    long long f;
}tree[4000006];
inline void build(long long i,long long x,long long y){
    tree[i].l=x,tree[i].r=y;
    if(x==y){
        tree[i].sum=rd();
        return ;
    }
    long long mid=(x+y)>>1;
    build(i<<1,x,mid);
    build(i<<1|1,mid+1,y);
    tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;
    return ;
}
inline void pushdown(long long i){
    if(!tree[i].f) return ;
    long long h=tree[i].f;
    tree[i<<1].f+=h,tree[i<<1|1].f+=h;
    tree[i<<1].sum+=(tree[i<<1].r-tree[i<<1].l+1)*h;
    tree[i<<1|1].sum+=(tree[i<<1|1].r-tree[i<<1|1].l+1)*h;
    tree[i].f=0;
    return ;
}
inline void add(long long i,long long x,long long y,long long z){
    if(tree[i].l>=x&&tree[i].r<=y){
        tree[i].f+=z;
        tree[i].sum+=(tree[i].r-tree[i].l+1)*z;
        return ;
    }
    pushdown(i);
    if(tree[i<<1].r>=x) add(i<<1,x,y,z);
    if(tree[i<<1|1].l<=y) add(i<<1|1,x,y,z);
    tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;
    return ;
}
inline long long solve(long long i,long long x,long long y){
    if(tree[i].l>=x&&tree[i].r<=y) return tree[i].sum;
    long long ans=0;
    pushdown(i);
    if(tree[i<<1].r>=x) ans+=solve(i<<1,x,y);
    if(tree[i<<1|1].l<=y) ans+=solve(i<<1|1,x,y);
    return ans;
}
char c[10006];
int main(){
    n=rd(),m=rd();
    build(1,1,n);
    for(long long i=1;i<=m;i++){
        scanf("%s",&c);
        if(c[0]=='C'){
            long long x=rd(),y=rd(),z=rd();
            add(1,x,y,z);
        }
        else{
            long long x=rd(),y=rd();
            write(solve(1,x,y)),puts("");
        }
    }
    return 0;
}