NOIp2012D1T2 国王游戏题解

国王游戏

洛谷P1080国王游戏

题解

这道题有个重要的性质：如果交换相邻两个大臣，获得金钱变化的有且只有这两个大臣。其余大臣得到的金钱不变。
我们考虑第 i i i个大臣和第 i + 1 i+1 i+1个大臣交换前后的变化。
交换前，这两个大臣的奖励是
1 B [ i ] × ∏ j = 0 i − 1 A [ j ] 与 1 B [ i + 1 ] × ∏ j = 0 i A [ j ] \frac {1}{B[i]}\times \prod_{j=0}^{i-1}A[j] 与 \frac {1}{B[i + 1]}\times \prod_{j=0}^{i}A[j] B[i]1×j=0∏i−1A[j]与B[i+1]1×j=0∏iA[j]
交换后，他们获得的奖励是：
1 B [ i + 1 ] × ∏ j = 0 i − 1 A [ j ] 与 A [ i + 1 ] B [ i ] × ∏ j = 0 i − 1 A [ j ] \frac 1{B[i + 1]}\times \prod_{j=0}^{i -1}A[j] 与\frac {A[i+1]}{B[i]}\times \prod_{j=0}^{i-1}A[j] B[i+1]1×j=0∏i−1A[j]与B[i]A[i+1]×j=0∏i−1A[j]
其他大臣的奖励都不会变，因此我们只要考虑上面两组数的最大值变化即可。
提取公因式 ∏ j = 0 j − 1 A [ j ] \prod _{j=0}^{j-1}A[j] ∏j=0j−1A[j]后，只需要比较下面两个式子的大小关系。
max ⁡ ( 1 B [ i ] , A [ i ] B [ i + 1 ] ) max ⁡ ( 1 B [ i + 1 ] , A [ i + 1 ] B [ i ] ) \max (\frac1{B[i]}, \frac {A[i]}{B[i+1]})\ \ \ \ \max(\frac {1}{B[i+1]}, \frac{A[i+1]}{B[i]}) max(B[i]1,B[i+1]A[i]) max(B[i+1]1,B[i]A[i+1])
同乘 B [ i ] × B [ i + 1 ] B[i]\times B[i+1] B[i]×B[i+1]
max ⁡ ( B [ i + 1 ] , A [ i ] × B [ i ] ) max ⁡ ( B [ i ] , A [ i + 1 ] × B [ i + 1 ] ) \max(B[i+1], A[i]\times B[i])\ \ \ \ \ \max(B[i],A[i +1]\times B[i+1]) max(B[i+1],A[i]×B[i]) max(B[i],A[i+1]×B[i+1])
易得 B [ i + 1 ] ≤ A [ i + 1 ] × B [ i + 1 ] B[i+1]\leq A[i+1]\times B[i+1] B[i+1]≤A[i+1]×B[i+1]和 A [ i ] × B [ i ] ≥ B [ i ] A[i]\times B[i]\geq B[i] A[i]×B[i]≥B[i]
于是 A [ i ] × B [ i ] ≤ A [ i + 1 ] × B [ i + 1 ] A[i]\times B[i]\leq A[i+1]\times B[i+1] A[i]×B[i]≤A[i+1]×B[i+1]的时候，交换前更优。所以只要按照 A [ i ] × B [ i ] A[i]\times B[i] A[i]×B[i]从小到大排序即可。
这道题统计答案要高精度。

code

#include<bits/stdc++.h>
using namespace std;
int read () {int num = 0; char c = ' '; int flag = 1;for (;c > '9' || c < '0'; c = getchar ())if (c == '-')flag = 0;for (;c >= '0' && c <= '9'; num = (num << 3) + (num << 1) + c - 48, c = getchar ());return flag ? num : - num;
}
int max (int a, int b) {return a > b ? a : b;
}
const int maxn = 1020;
struct stu {int l, r;
} x[maxn];
int mycmp (stu a, stu b) {return a.l * a.r < b.l * b.r;
}
int a, b, n;
void init () {n = read (), a = read (), b = read ();for (int i = 1;i <= n;i ++)x[i].l = read (), x[i].r = read ();
}
namespace GJD {int max (int a, int b) {return a > b ? a : b;} struct Bignum {int a[50000], n;};Bignum Read () {string s;cin >> s;Bignum x;x.n = s.length ();for (int i = 1;i <= x.n;i ++)x.a[x.n - i + 1] = s[i - 1] - 48;return x;}Bignum Print (Bignum p) {for (int i = p.n;i >= 1;i --)printf ("%d", p.a[i]);printf ("\n");}Bignum Clear (Bignum a) {a.n = 0;memset (a.a, 0, sizeof a.a);return a;}Bignum check (Bignum a) {for (int i = 1;i < a.n;i ++) {int m = a.a[i] / 10;if (m != 0) {a.a[i + 1] += m;a.a[i] %= 10;    }}while (a.a[a.n] >= 10) {a.n ++;a.a[a.n] = a.a[a.n - 1] / 10;a.a[a.n - 1] %= 10;}while (a.a[a.n] == 0 && a.n > 1) a.n --;return a;}Bignum Plus (Bignum a, Bignum b) {Bignum c; c = Clear (c);int N = max (a.n, b.n); c.n = N;for (int i = 1;i <= N;i ++)c.a[i] = a.a[i] + b.a[i];c = check (c);return c;}Bignum Mul (Bignum a, Bignum b) {Bignum c; c = Clear (c);int N = a.n + b.n - 1; c.n = N;for (int i = 1;i <= a.n;i ++)for (int j = 1;j <= b.n;j ++)c.a[i + j - 1] += a.a[i] * b.a[j];c = check (c);return c;}Bignum plus_small (Bignum a, int b) {a.a[1] += b;a = check (a);return a;}Bignum mul_small (Bignum a, int b) {for (int i = 1;i <= a.n;i ++)a.a[i] *= b;a = check (a);return a;}Bignum div_small (Bignum a, int b) {int now = 0;for (int i = a.n;i >= 1;i --) {now = now * 10 + a.a[i];a.a[i] = now / b;now %= b;}a = check (a);return a;}int bigger (Bignum a, Bignum b) {a = check (a); b = check (b);if (a.n > b.n) return 1;if (a.n < b.n) return 0;int N = a.n;for (int i = N;i >= 1;i --) {if (a.a[i] > b.a[i]) return 1;if (a.a[i] < b.a[i]) return 0;}return 2;}
} using namespace GJD;
void work () {Bignum now, ans;now = Clear (now); ans = Clear (ans);now.a[1] = a; now.n = 1;now = check (now);for (int i = 1;i <= n;i ++) {Bignum t = div_small (now, x[i].r);if (bigger (t, ans)) ans = t;now = mul_small (now, x[i].l);}Print (ans);
}
int main () {init ();sort (x + 1, x + 1 + n, mycmp);work ();return 0;
}