用差分实现区间操作
一维
给定一个长度为n的数列a，要求支持操作add(L,R,k)表示对a[L]~a[R]的每个数都加上k。并求修改后的序列a。
设原数组为a[i]a[i], 设数组d[i]=a[i]−a[i−1](a[0]=0)，则 a[i]=∑ij=1d[j]
更新操作update(s, t, d)把区间A[s]…A[t]都增加d，我们引入一个数组delta[i]，表示
A[i]…A[n]的共同增量，n是数组的大小。那么update操作可以转化为：
1）令delta[s] = delta[s] + d，表示将A[s]…A[n]同时增加d，但这样A[t+1]…A[n]就多加了d
2）再令delta[t+1] = delta[t+1] - d，表示将A[t+1]…A[n]同时减d
我们考虑用差分的做法。这里 需要一个辅助数组c，c用来记录某一个位置上的总改变量。c[i]表示的是i~n这些元素都加上c[i]这个数。我们对[L,R]区间进行加值操作，在c[L]处加一个k，在c[R+1]处就减去一个k。最后求序列的每个位置变成了多少，只需要求一下c数组的前缀和，然后和原数组按位相加就好。
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公式
w = (d + 1 + 2 * m + 3 * (m + 1) / 5 + y + (y >> 2) - y / 100 + y / 400) % 7;
小注
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w == 0 ? 7 : w ;
Description
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