codeforces --- Round #248 (Div. 2) B. Kuriyama Mirai's Stones
B. Kuriyama Mirai's Stones
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
- She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her
.
- Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her
.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
24
9
28
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
10
15
5
15
5
5
2
12
3
5
Please note that the answers to the questions may overflow 32-bit integer type.
【题目大意】
有一些石头排成一行,每个石头都有自己的权值,现在有两种询问:
一种是1开头的:询问原来的石头顺序下,第i到第j颗石头的的权值只和;
一种是2开头的,询问所有石头按照大小顺序排列以后,第i到第j颗石头的权值之和;
这题乍看很像树状数组,但是并不是树状数组,直接模拟即可.
#include<cstdio> #include<cstring> #include<algorithm> #define MAX 100100 using namespace std; __int64 m,a,b,c; __int64 n,i,sum; __int64 num1[MAX]; __int64 num2[MAX]; __int64 sum1[MAX]; __int64 sum2[MAX]; int main() { sum=0; scanf("%I64d",&n); for(i=0;i<n;i++) { scanf("%I64d",&num1[i]); sum+=num1[i]; sum1[i]=sum; } memcpy(num2,num1,sizeof(num1)); sort(num2,num2+n); sum=0; for(i=0;i<n;i++) { sum+=num2[i]; sum2[i]=sum; } scanf("%I64d",&m); while(m--) { scanf("%I64d%I64d%I64d",&a,&b,&c); if(a==1) printf("%I64d\n",sum1[c-1]-sum1[b-2]); else printf("%I64d\n",sum2[c-1]-sum2[b-2]); } return 0; }