正題

luogu 4774

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題目大意

有n條龍，第i條血量為$a_i$，回血量為$b_i$，殺死后掉落傷害為$D_i$的刀，初始有若干刀

殺第i條龍要用現有傷害比$a_i$小的刀中傷害最大的（如果沒有就用傷害最小的），對第i條龍造成x次傷害后，龍會連續回血，每次回$b_i$，若某一時刻龍的血量為0，那么該龍死亡

現在問你找到最小的x，使其滿足能殺死所有龍

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解題思路

對于每條龍所選刀，可以用multiset存現有刀，然后直接查詢符合的

設所選刀傷害為$d_i$，那么答案就是要求：

$$?(a_i?d_i\times x)\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{b}_i)$$

$$d_i\times x\equiv a_i(\mathrm{m}\mathrm{o}\mathrm{d}{b}_i)$$

該式子形如crt的式子

因為$b_i$不保證互質，所以求解要用excrt

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代碼

#include<set> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll __int128 #define N 100100 using namespace std; ll T, n, m, p, x, y, g, X, mx, a[N], mo[N], b[N]; multiset<ll>d; ll exgcd(ll a,ll b, ll &x, ll &y) {if(!b){x = 1;y = 0;return a;}ll k = exgcd(b, a % b, x, y);ll z = y;y = x - a / b * y;x = z;return k; } ll read() {char ch=getchar();ll ds=0,fs=1;while (ch<'0'||ch>'9') {if (ch=='-') fs=-1;ch=getchar();}while (ch>='0'&&ch<='9') ds=(ds<<3)+(ds<<1)+ch-48,ch=getchar();return ds*fs; } void writ(ll c) {if (c/10) writ(c/10); putchar(c%10+48); return;} void write(ll s) {s<0?putchar(45),writ(-s):writ(s); putchar(10);return;} int main() {T = read();while(T--){mx = 0;p = 0;d.clear();n = read();m = read();for (ll i = 1; i <= n; ++i)a[i] = read();for (ll i = 1; i <= n; ++i)mo[i] = read();for (ll i = 1; i <= n; ++i)b[i] = read();for (ll i = 1; i <= m; ++i)d.insert(read());for (ll i = 1; i <= n; ++i)//找刀{multiset<ll>::iterator it = d.upper_bound(a[i]);if (it == d.begin()) x = *it;else x = *--it;d.erase(it);d.insert(b[i]);b[i] = x;mx = max(mx, (a[i] - 1) / b[i] + 1);//最小刀數}m = 1;X = 0;for (ll i = 1; i <= n; ++i)//excrt{g = exgcd(b[i] * m, mo[i], x, y);if ((a[i] - X * b[i]) % g){p = 1;break;}x = x * ((a[i] - X * b[i]) / g) % (mo[i] / g);X = X + x * m;m = m * (mo[i] / g);X = (X % m + m) % m;//最小解}if (p){puts("-1");continue;}if (X < mx) X += (mx - X + m - 1) / m * m;//保證剩余血量>0write(X);}return 0; }

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