Codeforces Round #248 (Div. 2) B - Kuriyama Mirai's Stones

题意

给出一个n个数的序列。下面有m个指令

1指令：原序列从l到r的和

2指令：排序过的序列从l到r的和

思路

线段树可做….直接扫一遍记录sum数组也可以

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
#define ll long long
const int maxn = 100010;
int n;
int s[maxn];
int ss[maxn];
ll segTree[maxn<<2];
ll ssegTree[maxn<<2];
void build(int l,int r,int node)
{
    if(l == r) {
        segTree[node] = s[l];
        return;
    }
    build(l,(l+r)/2,node<<1);
    build((l+r)/2+1,r,(node<<1)+1);
    segTree[node] = segTree[node<<1]+segTree[(node<<1)+1];
}
void buildd(int l,int r,int node)
{
    if(l == r) {
        ssegTree[node] = ss[l];
        return;
    }
    buildd(l,(l+r)/2,node<<1);
    buildd((l+r)/2+1,r,(node<<1)+1);
    ssegTree[node] = ssegTree[node<<1]+ssegTree[(node<<1)+1];
}
ll query(int a,int b,int l,int r,int node)
{
    if(r < a || l > b) return 0;
    if(a <= l && r <= b) return segTree[node];
    return query(a,b,l,(l+r)/2,node<<1)+query(a,b,(l+r)/2+1,r,(node<<1)+1);
}
ll qquery(int a,int b,int l,int r,int node)
{
    if(r < a || l > b) return 0;
    if(a <= l && r <= b) return ssegTree[node];
    return qquery(a,b,l,(l+r)/2,node<<1)+qquery(a,b,(l+r)/2+1,r,(node<<1)+1);
}
int main()
{
    scanf("%d",&n);
    for(int i = 1 ; i <= n ; i ++) {
        scanf("%d",&s[i]);
        ss[i] = s[i];
    }
    sort(ss+1,ss+n+1);
    build(1,n,1);
    buildd(1,n,1);
    int m;
    scanf("%d",&m);
    while(m--) {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        if(a == 1) printf("%I64d\n",query(b,c,1,n,1));
        else printf("%I64d\n",qquery(b,c,1,n,1));
    }
    return 0;
}

Codeforces Round #248 (Div. 2) B - Kuriyama Mirai's Stones