UVa 11300 - Spreading the Wealth

来源：互联网发布：淘宝格子铺编辑：程序博客网时间：2024/05/16 05:24

这里写图片描述

分析：
把每个人的个数表示出来，如第一个人：A1 - X1 + X2 = M
可得X2 =M - A1 + X1 = X1 - C1(令C1 = M - A1)
以此类推，
最后找到规律，转化为数轴上一个点到N个点之间距离的问题，
发现当x取得c的中位数时最小，累加距离得出答案。

#include <iostream>#include <sstream>#include <iomanip>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <bitset>#include <string>#include <numeric>#include <algorithm>#include <functional>#include <iterator>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <cctype>#include <complex>#include <ctime>typedef long long LL;const double pi = acos(-1.0);const long long mod = 1e9 + 7;using namespace std;LL a[1000005];LL c[1000005];int main(){    //freopen("int.txt","r",stdin);    //freopen("out.txt","w",stdout);    int N;    while(scanf("%d",&N) == 1)    {        LL sum = 0;        for(int i = 1;i <= N;i++)        {            scanf("%I64d",&a[i]);            sum += a[i];        }        LL M = sum / N;        c[0] = 0;        for(int i = 1;i < N;i++)            c[i] = c[i - 1] + a[i] - M;        sort(c,c + N);        LL x1 = c[N / 2];        LL ans = 0;        for(int i = 0;i < N;i++)            ans += abs(x1 - c[i]);        printf("%I64d\n",ans);    }    return 0;}

//a数组其实可以不要#include <iostream>#include <sstream>#include <iomanip>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <bitset>#include <string>#include <numeric>#include <algorithm>#include <functional>#include <iterator>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <cctype>#include <complex>#include <ctime>typedef long long LL;const double pi = acos(-1.0);const long long mod = 1e9 + 7;using namespace std;LL c[1000005];int main(){    //freopen("int.txt","r",stdin);    //freopen("out.txt","w",stdout);    int N;    LL M,sum;    while(scanf("%d",&N) == 1)    {        sum = 0;        LL a;        c[0] = 0;        for(int i = 1;i <= N;i++)        {            scanf("%lld",&a);            sum += a;            c[i] = sum;        }        M = sum / N;        for(int i = 1;i < N;i++)            c[i] -= M * i;        sort(c,c + N);        LL x1 = c[N / 2],ans = 0;        for(int i = 0;i < N;i++)            ans += abs(x1 - c[i]);        printf("%lld\n",ans);    }    return 0;}

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