动态规划，而已！ CodeForces 433B - 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:

1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her .
2. 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.

Input

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.

Output

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.

Sample test(s)

input

6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6

output

24
9
28

input

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

output

10
15
5
15
5
5
2
12
3
5

Note

Please note that the answers to the questions may overflow 32-bit integer type.

题目说给一串数字，然后给指令1或2。输入l。r求第l到r的和。讲具体点：先输入一个n，代表有多少个元素，然后输入n个元素，然后输入一个q代表有多少次指令，1的指令就是直接将a[l]一直加加到a[r]，然后输出和，2的指令是先将a[]排序，sort即可了。然后和上面一样求和。思路非常easy，哈哈。看到这种B题非常开心吧，普通写法for循环一个个加的话。写完你就发现TLE了。并且TLE的非常开心啊！！！

！

！！

都说了这是动态规划啊。！！

！！！

咱们这样存储：每一个元素存储的是前i个元素的和。这样a[l]一直到a[r]的表达式就为a[r]-a[l-1];

好好理解下，对吧？。

动态规划就是拿空间换时间的算法，在执行过程中会产生大量中间数据进行抉择。每个状态始终影响下一步的状态！！！

！

！

！

嗯。贴代码时间：

#include
#include
#include
#include
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；

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