【随机算法】poj 2576/zoj 1880

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这题可用DP做，可是本人太菜，不懂，只会投机取巧^_^，所以用随机来AC，的确有点随机，个人觉关键是找准随机次数，例如这题，随机循环5w次刚好，这就难免贡献几次wa咯~

POJ 47MS / ZOJ 10MS。。。不错不错

#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <string.h>#include <deque>#include <stack>#include <algorithm>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>#include <time.h>using namespace std;int lowbit(int t){return t&(-t);}int countbit(int t){return (t==0)?0:(1+countbit(t&(t-1)));}int gcd(int a,int b){return (b==0)?a:gcd(b,a%b);}#define LL long long#define PI acos(-1.0)#define N  110#define MAX INT_MAX#define MIN INT_MIN#define eps 1e-8#define FRE freopen("a.txt","r",stdin)int a[N];int main(){    int n;    int i,j,k;    srand((unsigned int)time(0));    while(scanf("%d",&n)!=EOF){    int mid=n/2;    int ans1=0,ans2=0;    for(i=1;i<=n;i++){    scanf("%d",&a[i]);    if(i<=mid)ans1+=a[i];    else ans2+=a[i];    }    if(n==1){printf("%d %d\n",a[1],a[1]);continue;}    int minm=abs(ans1-ans2);    for(i=1;i<=50000;i++){        int l=(rand()%mid) + 1;        int r=(rand()%(n-mid)) + mid+1;        int t1=ans1-a[l]+a[r];        int t2=ans2-a[r]+a[l];        if(abs(t1-t2)<minm){            minm=abs(t1-t2);            ans1=t1;            ans2=t2;            int tmp=a[l];            a[l]=a[r];            a[r]=tmp;        }    }    if(ans1>ans2)swap(ans1,ans2);    printf("%d %d\n",ans1,ans2);    }    return 0;}