uva 11076

来源：互联网发布：量亿数据编辑：程序博客网时间：2024/06/06 07:14

计算出每一位上数字i会出现的次数累加

#include <cstdio>#include <cstdlib>#include <cmath>#include <map>#include <set>#include <stack>#include <vector>#include <sstream>#include <string>#include <cstring>#include <algorithm>#include <iostream>#define maxn 1010#define INF 0x7fffffff#define inf 10000000#define MOD 1000000007#define ull unsigned long long#define ll long longusing namespace std;int a[10], b[10], sum, f[13];void init(){    memset(a, 0, sizeof(a));    memset(b, 0, sizeof(b));    sum = 0;    f[0] = f[1] = 1;    for(int i = 2; i < 13; ++ i)        f[i] = f[i-1]*i;}int main(){    int n, r;    while(scanf("%d", &n) == 1 && n)    {        init();        for(int i = 0; i < n; ++ i) {scanf("%d", &r); a[r] ++; sum ++;}        for(int i = 1; i < 10; ++ i)            if(a[i]){                b[i] = f[sum-1];                for(int j = 0; j < 10; ++ j)                    if(a[j])                        b[i] /= i == j ? f[a[j]-1]:f[a[j]];            }        ll ans = 0;        for(int i = 0; i < n; ++ i)        {            ans *= 10;            for(int j = 0; j < 10; ++ j)                ans += j*b[j];        }        printf("%lld\n", ans);    }    return 0;}

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