hdu 2023wrong answer！

Posted 2023-05-02

#include<iostream>
using namespace std;
int main()

int n,m,g=0;
bool t=true;
float s;
cin>>n>>m;
float *a=new float[n*m];
float *b=new float[n];
float *c=new float[m];
for(int i=0;i<n*m;i++)
cin>>a[i];
for(i=0,s=0;i<m;i++)

for(int j=i;j<n*m;j+=m)
s+=a[j];
c[i]=s/m;
s=0;

for(i=0,s=0;i<n;i++)

for(int j=0;j<m;j++)

s+=a[i*m+j];
if(a[i*m+j]>c[j])
t=false;

if(t)
g++;
else
t=true;
b[i]=s/m;
s=0;

delete []a;
for(i=0;i<n;i++)

printf("%.2f",b[i]);
if(i==n-1)
cout<<endl;
else
cout<<" ";

delete []b;
for(i=0;i<m;i++)

printf("%.2f",c[i]);
if(i==m-1)
cout<<endl;
else
cout<<" ";

delete []c;
cout<<g<<endl;
return 0;

参考技术A 我AC的代码 C的。。你看下
#include"stdio.h"
int main()

int n,m;
int num[60][6];
double ave1[60],ave2[6];
int i,j;
int z;
while(scanf("%d%d",&n,&m)!=EOF)

for(i=0;i<n;i++)

ave1[i]=0;
for(j=0;j<m;j++)

scanf("%d",&num[i][j]);
ave1[i]+=num[i][j];

ave1[i]/=m;

for(i=0;i<n-1;i++)
printf("%0.2lf ",ave1[i]);
printf("%0.2lf\n",ave1[n-1]);
for(j=0;j<m;j++)

ave2[j]=0;
for(i=0;i<n;i++)
ave2[j]+=num[i][j];
ave2[j]/=n;
printf("%0.2lf",ave2[j]);
if(j!=m-1) printf(" ");

printf("\n");
for(i=0,z=0;i<n;i++)

for(j=0;j<m;j++)

if(num[i][j]<ave2[j])
break;

if(j==m)
z++;

printf("%d\n",z);
printf("\n");

return 0;
参考技术B 1、这题是多case~
你的程序只能处理单组数据，但实际上题目是多组数据连续写在一起的~
2、c[i]=s/m;
这里算平均成绩应该是除以人数嘛。。。改成c[i]=s/n;
3、if(a[i*m+j]>c[j])
你的意思是若有一个小于则不符合。但是，这个小于的方向错了。。。- -
4、输出格式改成0.2f试试。。。

hdu 3038 How Many Answers Are Wrong (带权并查集)

How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14876    Accepted Submission(s): 5237

Problem Description

TT and FF are ... friends. Uh... very very good friends -________-b

FF is a bad boy, he is always wooing TT to play the following game with him. This is a very humdrum game. To begin with, TT should write down a sequence of integers-_-!!(bored).

Then, FF can choose a continuous subsequence from it(for example the subsequence from the third to the fifth integer inclusively). After that, FF will ask TT what the sum of the subsequence he chose is. The next, TT will answer FF‘s question. Then, FF can redo this process. In the end, FF must work out the entire sequence of integers.

Boring~~Boring~~a very very boring game!!! TT doesn‘t want to play with FF at all. To punish FF, she often tells FF the wrong answers on purpose.

The bad boy is not a fool man. FF detects some answers are incompatible. Of course, these contradictions make it difficult to calculate the sequence.

However, TT is a nice and lovely girl. She doesn‘t have the heart to be hard on FF. To save time, she guarantees that the answers are all right if there is no logical mistakes indeed.

What‘s more, if FF finds an answer to be wrong, he will ignore it when judging next answers.

But there will be so many questions that poor FF can‘t make sure whether the current answer is right or wrong in a moment. So he decides to write a program to help him with this matter. The program will receive a series of questions from FF together with the answers FF has received from TT. The aim of this program is to find how many answers are wrong. Only by ignoring the wrong answers can FF work out the entire sequence of integers. Poor FF has no time to do this job. And now he is asking for your help~(Why asking trouble for himself~~Bad boy)

 

 

Input

Line 1: Two integers, N and M (1 <= N <= 200000, 1 <= M <= 40000). Means TT wrote N integers and FF asked her M questions.

Line 2..M+1: Line i+1 contains three integer: Ai, Bi and Si. Means TT answered FF that the sum from Ai to Bi is Si. It‘s guaranteed that 0 < Ai <= Bi <= N.

You can assume that any sum of subsequence is fit in 32-bit integer.

 

 

Output

A single line with a integer denotes how many answers are wrong.

 

 

Sample Input

10 5 1 10 100 7 10 28 1 3 32 4 6 41 6 6 1

 

 

Sample Output

1

 

 

Source

2009 Multi-University Training Contest 13 - Host by HIT

 

 

题目路径：

http://acm.hdu.edu.cn/showproblem.php?pid=3038

 

 

题意大概就是：一个数列，按顺序给你若干个区间的和，若某个区间的和与之前的冲突，就忽略，问你一共有多少个冲突的和。

 

带权并查集的一道例题。

 

带权并查集，其实就是说一个集合中的元素互相之间可以确定一种关系。初学时接触的并查集，其实就是说一个集合中的元素是一种“等价”关系，是带权并查集的一个特例。一个集合中的元素互相之间有关系，这便是（带权）并查集的本质。这点很重要，有必要理解下。

 

至于这道题，每个元素其实存放0...i的和(i的范围是1...n)以及与其父亲节点和的差值。find和union操作与简单并查集略有不同，主要是要更新两个值了。代码中只写了find（finding）操作，union操作在main函数中实现的。这题还有一个点，[a,b]的和就等于[0,b]-[0,a-1]的和（是a-1,而不是a）。

 

一些实现的细节还是要靠自己写，体会体会了。

 

还有一道更经典的例题，poj上的《食物链》。大家可以看一看，如果不会写的话（其实和这题差不多^_^），网上题解还是很多的。

 

 

#include <cstdio>

using namespace std;

const int maxn=200000;

struct
{
    int fa;
    int rel;//保存和其父亲节点前缀和的差值
}father[maxn+10];

int finding(int x)
{
    if(father[x].fa==x)
        return x;
    int tmp=father[x].fa;
    father[x].fa=finding(tmp);
    father[x].rel=father[x].rel+father[tmp].rel;//更新关系
    return father[x].fa;
}

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<=n;i++)//0..n而不是1..n(尽管点的编号是1..n)
        {
            father[i].fa=i;
            father[i].rel=0;
        }
        int ans=0;
        while(m--)
        {
            int a,b,s;
            scanf("%d%d%d",&a,&b,&s);
            a--;//这点文中已经有解释了

            int aroot=finding(a);
            int broot=finding(b);
            if(aroot==broot)
            {
                if(father[b].rel-father[a].rel!=-s)
                ans++;
            }
            else
            {
                father[aroot].fa=broot;
                father[aroot].rel=-father[a].rel+father[b].rel+s;
            }
            //printf("%d
",ans);
        }
        printf("%d
",ans);
    }
    return 0;
}

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