首页 > ACM题库 > HDU-杭电 > HDU 1060 Leftmost Digit[解题报告] C++
2013
11-26

HDU 1060 Leftmost Digit[解题报告] C++

Leftmost Digit

问题描述 :

Given a positive integer N, you should output the leftmost digit of N^N.

输入:

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

输出:

For each test case, you should output the leftmost digit of N^N.

样例输入:

2
3
4

样例输出:

2
2

Hint
In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2. In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1060

题意:n的n次方最左边的数字是多少。

mark:还是取对数那招,log10(n^n)的小数部分决定了最左边的数字。

代码:

# include <stdio.h>
 # include <math.h>
 
 
 int calc(long long n)
 {
     double ans = n * log10(n) ;
     ans -= (long long)ans ;
     return (int)pow(10,ans) ;
 }
 
 
 int main ()
 {
     int T, n ;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n) ;
         printf("%d\n",calc(n)) ;
     }
     return 0 ;
 }

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  2. #include <cstdio>
    #include <algorithm>

    struct LWPair{
    int l,w;
    };

    int main() {
    //freopen("input.txt","r",stdin);
    const int MAXSIZE=5000, MAXVAL=10000;
    LWPair sticks[MAXSIZE];
    int store[MAXSIZE];
    int ncase, nstick, length,width, tmp, time, i,j;
    if(scanf("%d",&ncase)!=1) return -1;
    while(ncase– && scanf("%d",&nstick)==1) {
    for(i=0;i<nstick;++i) scanf("%d%d",&sticks .l,&sticks .w);
    std::sort(sticks,sticks+nstick,[](const LWPair &lhs, const LWPair &rhs) { return lhs.l>rhs.l || lhs.l==rhs.l && lhs.w>rhs.w; });
    for(time=-1,i=0;i<nstick;++i) {
    tmp=sticks .w;
    for(j=time;j>=0 && store >=tmp;–j) ; // search from right to left
    if(j==time) { store[++time]=tmp; }
    else { store[j+1]=tmp; }
    }
    printf("%dn",time+1);
    }
    return 0;
    }