#1454. 【NOIP模拟（第二套）】B. 数连通块

https://sjzezoj.com/submission/54011

S2OJ/1454/1454.cpp

@@ -0,0 +1,85 @@
+#include <iostream>
+#include <vector>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 1e5 + 5;
+const int mod = 1e9 + 7;
+
+int n, m, q, p[N], s[N], c[N], ans[N];
+std::vector<int> g[N];
+std::vector<std::pair<int, int>> qs[N];
+
+int binpow(int a, int b) {
+    int res = 1;
+    a %= mod;
+
+    while (b) {
+        if (b & 1) res = static_cast<long long>(res) * a % mod;
+        a = static_cast<long long>(a) * a % mod;
+        b >>= 1;
+    }
+
+    return res;
+}
+
+inline int lowbit(int x) {
+    return x & -x;
+}
+
+void add(int x, int y) {
+    for (; x; x -= lowbit(x)) c[x] = (c[x] + y) % mod;
+}
+
+int sum(int x) {
+    int res = 0;
+    for (; x <= n; x += lowbit(x)) res = (res + c[x]) % mod;
+    return res;
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    cin >> n >> m >> q;
+
+    for (int i = 1, a, b; i <= n; i++) {
+        cin >> a >> b;
+
+        p[i] = static_cast<long long>(a) * binpow(b, mod - 2) % mod;
+        s[i] = (s[i - 1] + p[i]) % mod;
+    }
+
+    for (int i = 1, u, v; i <= m; i++) {
+        cin >> u >> v;
+
+        if (u > v) std::swap(u, v);
+
+        g[v].push_back(u);
+    }
+
+    for (int i = 1, l, r; i <= q; i++) {
+        cin >> l >> r;
+
+        ans[i] = (s[r] - s[l - 1]) % mod;
+        qs[r].emplace_back(i, l);
+    }
+
+    for (int i = 1; i <= n; i++) {
+        for (int v : g[i]) {
+            add(v, static_cast<long long>(p[i]) * p[v] % mod);
+        }
+
+        for (auto t : qs[i]) {
+            ans[t.first] = (ans[t.first] - sum(t.second)) % mod;
+        }
+    }
+
+    for (int i = 1; i <= q; i++) {
+        cout << (static_cast<long long>(ans[i]) % mod + mod) % mod << endl;
+    }
+
+    return 0;
+}