#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define st first
#define nd second
typedef long long ll;
typedef long double ld;
const ll I = 1'000'000'000'000'000'000LL;
const int II = 2'000'000'000;
const ll M = 1'000'000'007LL;
pair<int, int> operator*(const pair<int, int> &a, const pair<int, int> &b)
{
return pair{((ll)a.st * (ll)b.st) % M, ((ll)a.nd * (ll)b.st + (ll)b.nd) % M};
}
pair<int, int> &operator*=(pair<int, int> &a, const pair<int, int> &b)
{
return (a = pair{((ll)a.st * (ll)b.st) % M, ((ll)a.nd * (ll)b.st + (ll)b.nd) % M});
}
int operator*(const int &a, const pair<int, int> &b)
{
return ((ll)a * (ll)b.st + (ll)b.nd) % M;
}
int &operator*=(int &a, const pair<int, int> &b)
{
return (a = ((ll)a * (ll)b.st + (ll)b.nd) % M);
}
const int N = 1<<19;
pair<int, int> drz[2 * N];
pair<int, int> tab[N];
int nxt[N]; ll sum[N];
pair<int, int> Query(int a, int b)
{
a += N - 1; b += N + 1;
pair<int, int> res1 = pair{1, 0}, res2 = pair{1, 0};
while(a / 2 != b / 2)
{
if(a % 2 == 0)
res1 *= drz[a + 1];
if(b % 2 == 1)
res2 = (drz[b - 1]) * res2;
a /= 2; b /= 2;
}
return res1 * res2;
}
void Solve()
{
int n, q, a, b;
cin >> n >> q;
for(int i = 1; i <= n; ++i)
{
cin >> tab[i].nd >> tab[i].st;
if(tab[i].st > 1)
drz[i + N] = pair{tab[i].st, 0};
else
drz[i + N] = pair{1, tab[i].nd};
sum[i] = sum[i - 1] + tab[i].nd;
}
for(int v = N - 1; v >= 1; --v)
drz[v] = drz[v * 2] * drz[v * 2 + 1];
nxt[n + 1] = n + 1;
for(int i = n; i >= 1; --i)
{
if(tab[i].st >= 2)
nxt[i] = i;
else
nxt[i] = nxt[i + 1];
}
for(int i = 1; i <= q; ++i)
{
ll ans = 0LL;
int tmp;
cin >> tmp >> a >> b;
ans = tmp;
++a;
while(a <= b && ans < M)
{
ans += sum[min(nxt[a] - 1, b)] - sum[a - 1];
a = nxt[a];
if(ans < M && a <= b)
{
ans = max(ans * (ll)tab[a].st, ans + (ll)tab[a].nd);
++a;
}
}
tmp = ans % M;
pair<int, int> d = pair{1, 0};
if(a <= b)
d = Query(a, b);
tmp *= d;
cout << tmp << "\n";
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
//int t; cin >> t;
//while(t--)
Solve();
return 0;
}
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 | #include <bits/stdc++.h> using namespace std; #define pb push_back #define st first #define nd second typedef long long ll; typedef long double ld; const ll I = 1'000'000'000'000'000'000LL; const int II = 2'000'000'000; const ll M = 1'000'000'007LL; pair<int, int> operator*(const pair<int, int> &a, const pair<int, int> &b) { return pair{((ll)a.st * (ll)b.st) % M, ((ll)a.nd * (ll)b.st + (ll)b.nd) % M}; } pair<int, int> &operator*=(pair<int, int> &a, const pair<int, int> &b) { return (a = pair{((ll)a.st * (ll)b.st) % M, ((ll)a.nd * (ll)b.st + (ll)b.nd) % M}); } int operator*(const int &a, const pair<int, int> &b) { return ((ll)a * (ll)b.st + (ll)b.nd) % M; } int &operator*=(int &a, const pair<int, int> &b) { return (a = ((ll)a * (ll)b.st + (ll)b.nd) % M); } const int N = 1<<19; pair<int, int> drz[2 * N]; pair<int, int> tab[N]; int nxt[N]; ll sum[N]; pair<int, int> Query(int a, int b) { a += N - 1; b += N + 1; pair<int, int> res1 = pair{1, 0}, res2 = pair{1, 0}; while(a / 2 != b / 2) { if(a % 2 == 0) res1 *= drz[a + 1]; if(b % 2 == 1) res2 = (drz[b - 1]) * res2; a /= 2; b /= 2; } return res1 * res2; } void Solve() { int n, q, a, b; cin >> n >> q; for(int i = 1; i <= n; ++i) { cin >> tab[i].nd >> tab[i].st; if(tab[i].st > 1) drz[i + N] = pair{tab[i].st, 0}; else drz[i + N] = pair{1, tab[i].nd}; sum[i] = sum[i - 1] + tab[i].nd; } for(int v = N - 1; v >= 1; --v) drz[v] = drz[v * 2] * drz[v * 2 + 1]; nxt[n + 1] = n + 1; for(int i = n; i >= 1; --i) { if(tab[i].st >= 2) nxt[i] = i; else nxt[i] = nxt[i + 1]; } for(int i = 1; i <= q; ++i) { ll ans = 0LL; int tmp; cin >> tmp >> a >> b; ans = tmp; ++a; while(a <= b && ans < M) { ans += sum[min(nxt[a] - 1, b)] - sum[a - 1]; a = nxt[a]; if(ans < M && a <= b) { ans = max(ans * (ll)tab[a].st, ans + (ll)tab[a].nd); ++a; } } tmp = ans % M; pair<int, int> d = pair{1, 0}; if(a <= b) d = Query(a, b); tmp *= d; cout << tmp << "\n"; } } int main() { ios_base::sync_with_stdio(false); cin.tie(nullptr); //int t; cin >> t; //while(t--) Solve(); return 0; } |
English