Codevs1198 国王游戏

• 思路：贪心。由题意可知，交换相邻两个人对于其他人并无影响。设这两个人按顺序分别为{a1，b1}，{a2，b2}，交换前是max{sum*1/b1,sum*a1/b2}，交换后是max{sum*1/b2,sum*a2/b1}，而分母为1的项一定不会被最大值考虑到，则需要考虑max{sum*a2/b1,sum*a1/b2}，即max{a2/b1,a1/b2}。如果不交换更优，则有a1/b2<a2/b1为关键字进行从小到大排序，会得到最优解。最后只需扫描一遍序列就行了。由于乘积可能很大，所以需要使用高精度计算。

• 代码如下:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1005;struct node{        int a,b;    node(int a=0,int b=0):a(a),b(b){}    bool operator < (const node &x) const    {        return a*b<x.a*x.b;    }};struct bignum{    int len,a[5000];    bignum()    {                len=1;        memset(a,0,sizeof(a));    }    bignum operator * (int num)    {        len++;        int jinwei=0;        for (int i=1;i<=len;++i)        {            a[i]=a[i]*num+jinwei;            jinwei=a[i]/100000;            a[i]%=100000;        }        if (a[len]==0)          len--;        return *this;    }    bool operator < (const bignum &b) const    {        if (b.len!=len)          return len<b.len ? 1:0;        for (int i=len;i>=1;--i)          if (a[i]!=b.a[i])            return a[i]<b.a[i] ? 1:0;        return 0;    }    bignum operator / (int num)    {        bignum c=*this;        int jiewei=0;        for (int i=len;i>=1;--i)        {            jiewei=a[i]+100000*jiewei;            c.a[i]=jiewei/num;            jiewei%=num;        }        while (c.a[c.len]==0)          c.len--;        return c;    }    bignum operator *= (int num)    {        *this=*this*num;        return *this;    } };void print(bignum x){    printf("%d",x.a[x.len]);    for (int i=x.len-1;i>=1;--i)      printf("%05d",x.a[i]);}int n;node man[maxn];int main(){    scanf("%d",&n);    scanf("%d%d",&man[1].a,&man[1].b);    for (int i=2;i<=n+1;++i)    {        int a,b;        scanf("%d%d",&a,&b);        man[i]=node(a,b);    }    sort(man+2,man+n+2);    bignum ans,sum;    sum.a[1]=man[1].a;    for (int i=2;i<=n+1;++i)    {        ans=max(sum/man[i].b,ans);        sum*=man[i].a;    }    print(ans);    return 0;}