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

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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 longconst 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;}

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