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  • POJ 3077-Rounders(水题乱搞)

    Rounders
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 7697   Accepted: 4984

    Description

    For a given number, if greater than ten, round it to the nearest ten, then (if that result is greater than 100) take the result and round it to the nearest hundred, then (if that result is greater than 1000) take that number and round it to the nearest thousand, and so on ...

    Input

    Input to this problem will begin with a line containing a single integer n indicating the number of integers to round. The next n lines each contain a single integer x (0 <= x <= 99999999).

    Output

    For each integer in the input, display the rounded integer on its own line. 

    Note: Round up on fives.

    Sample Input

    9
    15
    14
    4
    5
    99
    12345678
    44444445
    1445
    446

    Sample Output

    20
    10
    4
    5
    100
    10000000
    50000000
    2000
    500
    题意:给一个数字。然后从最后一位開始进位,满5进1。小于5变成0。比方 12345 -> 12350->12400->12000->10000;
    暴力乱搞即可了。

    #include <algorithm>
    #include <iostream>
    #include <cstring>
    #include <cstdlib>
    #include <string>
    #include <cctype>
    #include <vector>
    #include <cstdio>
    #include <cmath>
    #include <deque>
    #include <stack>
    #include <map>
    #include <set>
    #define ll long long
    #define maxn 116
    #define pp pair<int,int>
    #define INF 0x3f3f3f3f
    #define max(x,y) ( ((x) > (y)) ? (x) : (y) )
    #define min(x,y) ( ((x) > (y)) ?

    (y) : (x) ) using namespace std; int x; char s[15]; void solve() { sprintf(s,"%d",x); for(int i=strlen(s)-1;i>0;i--) { if(s[i]>='5') { s[i]='0'; s[i-1]++; } else s[i]='0'; } if(s[0]>'9'){s[0]='0';printf("1");} puts(s); } int main() { int T; scanf("%d",&T); while(T--) { scanf("%d",&x); if(x<=10){printf("%d ",x);continue;} solve(); } return 0; }

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  • 原文地址:https://www.cnblogs.com/mfmdaoyou/p/6742786.html
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