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.
！
！！
！！！
！
！
！
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define maxn 100006
__int64 sum;
__int64 pp[maxn]={0},p[maxn]={0},liu[maxn],xp[maxn];
int main()
{
int i,j,k;
int t,n,m;
int l,r;
while(scanf("%d",&n)!=EOF)
{
p[0]=0;
for(i=1;i<=n;i++)
{
scanf("%I64d",&liu[i]);
xp[i]=liu[i];
p[i]=p[i-1]+liu[i];
}
xp[0]=0;
sort(xp,xp+n+1);
for(i=1;i<=n;i++)
pp[i]=pp[i-1]+xp[i];
scanf("%d",&t);
while(t--)
{
scanf("%d",&m);
if(m==1)
{
scanf("%d%d",&l,&r);
sum=p[r]-p[l-1];
printf("%I64d\n",sum);
}
else if(m==2)
{
scanf("%d%d",&l,&r);
sum=pp[r]-pp[l-1];
printf("%I64d\n",sum);
}
}
}
return 0;
}看出bug就讲吧，谢谢；
版权声明：本文博客原创文章，博客，未经同意，不得转载。
本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/4675296.html，如需转载请自行联系原作者
