Loading docs/ds/bit.md +2 −2 Original line number Diff line number Diff line Loading @@ -90,7 +90,7 @@ int getsum(int x) { // a[1]……a[x]的和 } ``` ## 区间加区间求和 ## 区间修改加区间求和 若维护序列 $a$ 的差分数组 $b$ ,此时我们对 $a$ 的一个前缀 $r$ 求和,即 $\sum_{i=1}^{r} a_i$ ,由差分数组定义得 $a_i=\sum_{j=1}^i b_j$ Loading Loading @@ -132,7 +132,7 @@ void add1(int l, int r, int v) { } long long getsum1(int l, int r) { return (r + 1ll) * (getsum(t1, r) - getsum(t1, l - 1)) - return (r + 1ll) * getsum(t1, r) - 1ll * l * getsum(t1, l - 1) - (getsum(t2, r) - getsum(t2, l - 1)); } ``` Loading Loading
docs/ds/bit.md +2 −2 Original line number Diff line number Diff line Loading @@ -90,7 +90,7 @@ int getsum(int x) { // a[1]……a[x]的和 } ``` ## 区间加区间求和 ## 区间修改加区间求和 若维护序列 $a$ 的差分数组 $b$ ,此时我们对 $a$ 的一个前缀 $r$ 求和,即 $\sum_{i=1}^{r} a_i$ ,由差分数组定义得 $a_i=\sum_{j=1}^i b_j$ Loading Loading @@ -132,7 +132,7 @@ void add1(int l, int r, int v) { } long long getsum1(int l, int r) { return (r + 1ll) * (getsum(t1, r) - getsum(t1, l - 1)) - return (r + 1ll) * getsum(t1, r) - 1ll * l * getsum(t1, l - 1) - (getsum(t2, r) - getsum(t2, l - 1)); } ``` Loading