2984. 数学手稿

Andrew-Malcom

include

using namespace std;
struct Node{
        int x,y,z;
}stu[1000001];
bool cmp(Node a,Node b)
{
        return a.x*10000+a.y*100+a.z<b.x*10000+b.y*100+b.z;
}
int main()
{
        int t,i;cin>>t;
        for(i=0;i<t;i++){
                int k,n;cin>>n>>k;
                int j,index=0;
                if(k-n<0||2*n-k<0){
                        cout<<"case #"<<i<<":\n";
                        cout<<-1<<endl;
                        continue;
                }
                for(j=k-2*n;j<=(k-n)/2;j++){
                        if(j>=0&&k-n-2*j>=0&&2*n-k+j>=0){
                                stu[index].x=j;
                                stu[index].y=k-n-2*j;
                                stu[index++].z=2*n-k+j;
                        }
                }
                sort(stu,stu+index,cmp);
                cout<<"case #"<<i<<":\n";
                for(j=0;j<index;j++){
                        cout<<stu[j].x<<" "<<stu[j].y<<" "<<stu[j].z<<endl;
                }
        }
}
LUSTRE

include

include

int main()
{
int T,Ti;
scanf(“%d”,&T);
for(Ti=0; Ti<T; Ti++)
{
int N,flag=0;
long long K,m,w,c;
scanf(“%d%lld”,&N,&K);
printf(“case #%d:\n”,Ti);
for(m=0; m<=N; m++)
for(w=0; w<=N; w++)
for(c=0; c<=N; c++)
if(m+w+c==N&&3m+2w+c==K)
{
printf(“%lld %lld %lld\n”,m,w,c);
flag=1;
}
if(flag==0)
printf(“-1\n”);
}
return 0;
}

aiden
#include <iostream>
using namespace std;
struct distribution
{
    int m;
    int w;
    int c;
};
int cmp(const void *a, const void *b)
{
    distribution *p1 = (distribution *)a;
    distribution *p2 = (distribution *)b;
    return (p1->m * 10000 + p1->w * 100 + p1->c) - (p2->m * 10000 + p2->w * 100 + p2->c);
}
int main()
{
    int t;
    cin >> t;
    for (int z = 0; z < t; z++)
    {
        int n, k;
        cin >> n >> k;
        distribution list[1000];
        int m, w, c;
        int cnt = 0;
        for (m = 0; (m <= k / 3) && m <= n; m++)
        {
            for (w = 0; (w <= k / 2) && (m + w <= n); w++)
            {
                for (c = 0; (c <= k) && (m + w + c <= n); c++)
                {
                    if ((3 * m + 2 * w + c == k) && (m + w + c == n))
                    {
                        list[cnt].m = m;
                        list[cnt].w = w;
                        list[cnt].c = c;
                        cnt++;
                    }
                }
            }
        }
        cout << "case #" << z << ":" << endl;
        if (cnt == 0)
        {
            cout << "-1" << endl;
            continue;
        }
        qsort(list, cnt, sizeof(distribution), cmp);
        for (auto i = 0; i < cnt; i++)
        {
            cout << list[i].m << ' ' << list[i].w << ' ' << list[i].c << endl;
        }
    }
    return 0;
}
一剑无痕雪满山

分享一个过了一周才弄出来的智障代码
_(:зゝ∠)_总是WA要哭了
然后发现是因为没有输出只有男人或者女人或者小孩的情况。。

include

include

include

int cmp1(const void a,const void b)
{ return (int)a-(int)b;}
void solve()
{
int N=0;long long int K=0;
scanf(“%d %I64d”,&N,&K);
int man;int woman;int children;int count=233;
int a[10000]={0},b[10000]={0},c[10000]={0};long long int sum[10000]={0};int i=0;int j;

 for (man=0;man<=N;man++)
    {

        for (woman=0;woman<=N;woman++)
         {

             for (children=0;children<=N;children++)
                 {

                     if((3*man+2*woman+children==K)&&(man+woman+children==N))
          {
            a[i]=man;b[i]=woman;c[i]=children;count=1;i++;
          } }}
    }

for (j=0;j<i;j++)
{sum[j]=a[j]10000+b[j]100+c[j];}
qsort(sum,i,sizeof(sum[0]),cmp1);
for (j=0;j<i;j++)
{

    printf("%d %d %d\n",a[j],b[j],c[j]);

}
if (N==0&&K==0) {printf(“0 0 0\n”);}
else if (count==233) {printf(“-1\n”);}

memset(a,0,sizeof(a));
}

int main()
{
int i,t;
scanf(“%d”,&t);
for (i=0;i<t;i++) {printf(“case #%d:\n”,i); solve(); }
return 0;
}

徐摆渡

两个方程三个未知数 可以用一个变量来表示另外两个变量 然后判断解的情况即可

#include <bits/stdc++.h>
using namespace std;

int T, cnt, m, w, c;
struct data
{
    int m;
    int w;
    int c;
    int ans;
}d[1000];

bool cmp(data x, data y)
{
    return x.ans<y.ans;
}

void solve(int n, int k)
{
    int max_=(3*n-k)/2, min_=2*n-k;
    int l=0;
    if(max_<0||max_<min_) cout<<"-1"<<endl;
    else if(min_<0){
        for(c=0;c<=max_;++c){
            d[l].m=k-2*n+c;
            d[l].w=3*n-k-2*c;
            d[l].c=c;
            d[l].ans=10000*m+100*w+c;
            l++;
        }
    }
    else {
        for(c=min_;c<=max_;++c){
            d[l].m=k-2*n+c;
            d[l].w=3*n-k-2*c;
            d[l].c=c;
            d[l].ans=10000*m+100*w+c;
            l++;
        }
    }
    sort(d,d+l,cmp);
    for(int i=0;i<l;++i)
        printf("%d %d %d\n",d[i].m, d[i].w, d[i].c);
}
int main()
{
    cin>>T;
    while(T--){
        int n, k;
        cin>>n>>k;
        printf("case #%d:\n",cnt++);
        solve(n,k);
    }
    return 0;
}
13627999316

屎山代码

徐摆渡

大一写的代码..现在看确实很冗长..

Master X

这题不要想任何骚操作…………………………
直接三重就行………………

kcisemedamuoy

分享一个过了一周才弄出来的智障代码
真的非常谢谢!如果没有提醒的话说不定会僵持很久

YZAZJL

include

using namespace std;

int main()
{
int n;
cin >> n;
int i;
for(i = 0; i < n; i++){
int total, money;
cin >> total >> money;
int m, w, c;
int cnt = 0;
cout << “case #” << i << “: ” << endl;
for(m = 0; m <= money / 3; m++){
for(w = 0; w <= total - m && w <= money / 2; w++){
for(c = 0; c <= total - m - w && c <= money; c++){
if(m + w + c == total && 3m + 2w + c == money){
cnt++;
cout << m << ” ” << w << ” ” << c << endl;
}
}
}
}
if(cnt == 0){
cout << -1 << endl;
}
}
return 0;
}

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