XTOJ1250Super Fast Fourier Transform

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链接：http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1250

题意：给定两个数组a,b，计算sigma(i,j) (int)sqrt(abs(a[i]-b[j]))，a,b数组的和都<=1e6。

分析：因为a数组的和<=1e6，所以不同的数字最多有1000个，同理b数组也是，然后只要将所有的数1e3*1e3统计一下就行了。

代码：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<cstdio>#include<vector>#include<string>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=100010;const int MAX=1000000100;const int mod=100000000;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=998244353;const int INF=1000000010;const double pi=acos(-1.0);typedef double db;typedef unsigned long long ull;int a[2][1010],b[2][1010],g[N*10];int main(){    int i,j,n,m,x,tota,totb;    ll ans;    while (scanf("%d%d", &n, &m)!=EOF) {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        tota=totb=0;        memset(g,0,sizeof(g));        for (i=1;i<=n;i++) {            scanf("%d", &x);g[x]++;        }        for (i=0;i<=1000000;i++)        if (g[i]) {            tota++;a[0][tota]=g[i];a[1][tota]=i;        }        memset(g,0,sizeof(g));        for (i=1;i<=m;i++) {            scanf("%d", &x);g[x]++;        }        for (i=0;i<=1000000;i++)        if (g[i]) {            totb++;b[0][totb]=g[i];b[1][totb]=i;        }        ans=0LL;        for (i=1;i<=tota;i++)            for (j=1;j<=totb;j++)            ans+=(ll)a[0][i]*b[0][j]*((ll)sqrt(abs(a[1][i]-b[1][j])));        printf("%lld\n", ans);    }    return 0;}

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