2014牡丹江网络预选赛E题（线段树）zoj3813

Alternating Sum

Time Limit: 2 Seconds Memory Limit: 65536 KB

There is a digit string S with infinite length. In addition, S is periodic and it can be formed by concatenating infinite repetitions of a base string P.
For example, if P = 3423537, then S = 3423537342353734235373423537...
Let's define the alternating sum on substrings of S. Assume Sl..r is a substring of S from index l to index r (all
indexes are 1-based), then the alternating sum of Sl..r is:

G(l, r) = Sl - Sl+1 + Sl+2 - ... + (-1)r-lSr
For example, S2..10 = 423537342, then G(2, 10) = 4 - 2 + 3 - 5 + 3 - 7 + 3 - 4 + 2 = -3.
Now, you are given the base string P and you have to do many operations. There are only two kinds of operations:

1 x d: set Px to d, d is a single digit.
2 l r: find the sum of G(i, j) that l <= i <= j <= r.

For each second operation, you should output the sum modulo 109 + 7.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
The first line contains a digit string P (1 <= length(P) <= 100000).
The second line contains an integer Q (1 <= Q <= 100000) indicating the number of operations. Each of the following Q lines is an operation in such format:

1 x d (1 <= x <= length(P), 0 <= d <= 9)
2 l r (1 <= l <= r <= 1018)

Output

For each "2 l r" operation, output an integer, indicating the sum modulo 109 + 7.

Sample Input

2
324242
4
2 1 1
2 1 4
1 3 7
2 3 4
324242
6
2 1 1
1 3 7
2 2 4
1 3 4
2 7 10
2 1 30

Sample Output

3
20
14
3
8
20
870

题意：RT

思路：感觉做这种题要很有耐心= =

首先不能被G（i,j）吓到，像我这种人就是比赛的时候一看到这个就被吓到了，果断放弃了

事实上赛后我再冷静的把一个区间里的SUM （G（i,j））写出来后发现有个很明显的规律

SUM（G（i,j））= p[i]*(j-i+l) + p[i+2]*(j-(i+2)+1) +......+p[j]*1 （1）

或SUM（G（i,j））= p[i]*(j-i+l) + p[i+2]*(j-(i+2)+1) +......+p[j-1]*2 （2）

如果区间长度为奇数就是（1）式，为偶数则为（2）式

看出这个规律以后要怎么搞呢，其实我觉得看出这个公式只是最基本的，后面的才叫恶心= =

每次要查询[l,r]区间的SUM值，l和r可以很大，显然这也是有规律的

不难想到一般求区间和可以转化为求SUM（G（1，r））-SUM（G（1，l-1））

这样就转化为了求前缀和，这样显然对于这种恶心题来讲会好处理很多

为了方便处理，将整个P序列直接复制一份到后面，现在可以用线段树维护整个2*P序列

可以看出求SUM的式子求和不是累加连续的数，而是隔一个数加一次

这样就需要将数的位置分奇偶分别求和，这两部分是不相干的

接下来就是求前缀和自己慢慢搞吧，已经没啥了，就推一下简单的求和公式而已

还是感觉这题有点恶心= =