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:

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.

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

input

output

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<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；