ZOJ 3813 Alternating Sum (牡丹江网络赛E题）

ZOJ 3813 Alternating Sum

题目链接

赛后补题中，这题真心恶心爆了

先推下公式，发现是隔一个位置，长度从最长每次减2，这样累加起来的和，然后就可以利用线段树维护，记录4个值，奇数和，偶数和，奇数答案和，偶数答案和，这样pushup的时候，对应要乘系数其实就是加上左边奇(偶)和乘上右边长度，线段树处理完，还有个问题就是查询可能横跨很多个区间，这样一来就要把区间进行分段，分成3段，然后和上面一样的方法合并即可，注意中间一段很长，不能一一去合并，可以推成等差数列，利用前n项和去搞

然后这题也是一直WA啊！！，一怒之下把所有可能爆longlong的地方全处理了，结果就过了- -

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define lson(x) ((x<<1) + 1)#define rson(x) ((x<<1) + 2)const long long N = 200005;typedef long long ll;const ll MOD = 1000000007;long long t, n;char str[N];struct Node {long long l, r;ll odd, even, sum1, sum2;ll size() {return r - l + 1;}} node[N * 4];void pushup(long long x) {if (node[lson(x)].size() % 2) {node[x].odd = (node[lson(x)].odd + node[rson(x)].even) % MOD;node[x].even = (node[lson(x)].even + node[rson(x)].odd) % MOD;node[x].sum1 = ((node[lson(x)].sum1 + node[rson(x)].size() * (node[lson(x)].odd % MOD) % MOD) % MOD + node[rson(x)].sum2) % MOD;node[x].sum2 = ((node[lson(x)].sum2 + node[rson(x)].size() * (node[lson(x)].even % MOD) % MOD) % MOD + node[rson(x)].sum1) % MOD;} else {node[x].odd = (node[lson(x)].odd + node[rson(x)].odd) % MOD;node[x].even = (node[lson(x)].even + node[rson(x)].even) % MOD;node[x].sum1 = ((node[lson(x)].sum1 + node[rson(x)].size() * (node[lson(x)].odd % MOD) % MOD) % MOD + node[rson(x)].sum1) % MOD;node[x].sum2 = ((node[lson(x)].sum2 + node[rson(x)].size() * (node[lson(x)].even % MOD) % MOD) % MOD + node[rson(x)].sum2) % MOD;}}void build(long long l, long long r, long long x = 0) {node[x].l = l; node[x].r = r;if (l == r) {node[x].odd = str[l] - '0';node[x].even = 0;node[x].sum1 = str[l] - '0';node[x].sum2 = 0;return;}long long mid = (l + r) / 2;build(l, mid, lson(x));build(mid + 1, r, rson(x));pushup(x);}void add(long long l, long long r, long long v, long long x = 0) {if (node[x].l >= l && node[x].r <= r) {node[x].odd = v;node[x].even = 0;node[x].sum1 = v;node[x].sum2 = 0;return;}long long mid = (node[x].l + node[x].r) / 2;if (l <= mid) add(l, r, v, lson(x));if (r > mid) add(l, r, v, rson(x));pushup(x);}Node get(long long l, long long r, long long tp, long long x = 0) {if (node[x].l >= l && node[x].r <= r)return node[x];long long mid = (node[x].l + node[x].r) / 2;if (l <= mid && r > mid) {Node ln = get(l, r, tp, lson(x));Node rn;Node ans;if (ln.size() % 2) {rn = get(l, r, !tp, rson(x));ans.l = ln.l;ans.r = rn.r;ans.odd = (ln.odd + rn.even) % MOD;ans.even = (ln.even + rn.odd) % MOD;ans.sum1 = ((ln.sum1 + rn.sum2) % MOD + rn.size() * (ln.odd % MOD) % MOD) % MOD;ans.sum2 = ((ln.sum2 + rn.sum1) % MOD + rn.size() * (ln.even % MOD) % MOD) % MOD;} else {rn = get(l, r, tp, rson(x));ans.l = ln.l;ans.r = rn.r;ans.odd = (ln.odd + rn.odd) % MOD;ans.even = (ln.even + rn.even) % MOD;ans.sum1 = ((ln.sum1 + rn.sum1) % MOD + rn.size() * (ln.odd % MOD) % MOD) % MOD;ans.sum2 = ((ln.sum2 + rn.sum2) % MOD + rn.size() * (ln.even % MOD) % MOD) % MOD;}return ans;} else if (l <= mid) return get(l, r, tp, lson(x));else if (r > mid) return get(l, r, tp, rson(x));}ll query(long long l, long long r, long long tp) {Node tmp = get(l, r, tp);if (tp == 0) return tmp.sum1;return tmp.sum2;}ll sum(ll a0, ll ti, ll an, ll d) {if (d == 0) return a0 % MOD * ti % MOD;ll ci = ti + 1;ll b = (a0 + an);if (b % 2 == 0)b /= 2;else if (ci % 2 == 0)ci /= 2;return b % MOD * (ci % MOD) % MOD;}ll solve(ll l, ll r) {ll s = l % n;ll e = r % n;if (s == 0) s = n;if (e == 0) e = n;ll ti = (r - e - (l + (n - s))) / n;if (ti < 0)return query(s, e, 0) % MOD;long long tp = 1;if ((n - s + 1) % 2) tp = 0;if (ti == 0) {Node ls = get((long long)s, n, 0);Node rs = get(1, (long long)e, tp);ll ans;if (tp) ans = (ls.sum1 + rs.sum1) % MOD;else ans = (ls.sum1 + rs.sum2) % MOD;ans = (ans + ls.odd * e) % MOD;return ans;}Node ls = get((long long)s, n, 0);Node mids = get(1, n, tp);Node rs = get(1, (long long)e, tp);ll ans;if (tp) ans = ((ls.sum1 + mids.sum1 % MOD * (ti % MOD) % MOD) % MOD + rs.sum1) % MOD;else ans = ((ls.sum1 + mids.sum2 % MOD * (ti % MOD) % MOD) % MOD +rs.sum2) % MOD;ll a0 = ls.odd;ll d;if (tp) d = mids.odd;else d = mids.even;ll an = a0 + (ti - 1) * d;ans = (ans + sum(a0, ti - 1, an, d) % MOD * n % MOD) % MOD;an += d;ans = (ans + an % MOD * e % MOD) % MOD;return ans;}int main() {scanf("%lld", &t);while (t--) {scanf("%s", str + 1);n = strlen(str + 1);for (long long i = 1; i <= n; i++)str[i + n] = str[i];n *= 2;build(1, n);long long q;scanf("%lld", &q);long long v, x, d;ll l, r;while (q--) {scanf("%lld", &v);if (v == 1) {scanf("%lld%lld", &x, &d);add(x, x, d);add(x + n / 2, x + n / 2, d);} else {scanf("%lld%lld", &l, &r);printf("%lld\n", solve(l, r));}}}return 0;}