E -- XJT Love Digits-----水题

E -- XJT Love Digits

Time Limit：1s Memory Limit：64MByte

Submissions：298Solved：88

DESCRIPTION

Define a function f(x) = the sum of digits of x, such as f(123) = 6. You will be given one sequence$a_1,a_2,a_3,...,a_n$, and q queries. Each query contains a number x, you need to output max number p among$a_1,...,a_x,a_{x-1}$,which satisfies f(p) = f($a_x$).

INPUT

The first line of input contains an integer$T(T\le 5)$ indicating the number of test cases.The first line of each case contains two interger$n(1\le n\le 100000)$ and $q(1\le q\le 100000)$.The second line contains n numbers: $a_1,a_2,...a_n(1\le a_i\le {10}^9)$.Following q lines, each line contains a integer $x(1\le x\le n)$.

OUTPUT

For each case, output "Case #X:" in a line where X is the case number, staring from 1. Then for each query, output one line. Each line contains one integer p. If there is no such p among$a_1,...,a_x$ which satisfies f(p) = f($a_x$), then just output -1.

SAMPLE INPUT

16 61 2 20 11 4 1654321

SAMPLE OUTPUT

Case #1:1-1202-1-1

题目链接：http://www.ifrog.cc/acm/problem/1083?contest=1010&no=4

玲珑杯的水题，没什么好说的。

代码：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;struct node{    int maxn;    int data;}xin[200000];int a[200];int bit(int n){    int x=n;    int ans=0;    while(x){        ans+=x%10;        x/=10;    }    return ans;}int main(){    int t;    scanf("%d",&t);    int cnt=0;    while(t--){        cnt++;        memset(a,-1,sizeof(a));        int n,m;        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++){            scanf("%d",&xin[i].data);            int h=bit(xin[i].data);            xin[i].maxn=a[h];            a[h]=max(a[h],xin[i].data);        }        printf("Case #%d:\n",cnt);        for(int i=0;i<m;i++){            int x;            scanf("%d",&x);            printf("%d\n",xin[x].maxn);        }    }    return 0;}