2016 CCPC Hangzhou Onsite A - ArcSoft's Office Rearrangement(HDU 5933) (贪心，模拟）

ArcSoft, Inc. is a leading global professional computer photography and computer vision technology company.

There are NN working blocks in ArcSoft company, which form a straight line. The CEO of ArcSoft thinks that every block should have equal number of employees, so he wants to re-arrange the current blocks into KK new blocks by the following two operations:

- merge two neighbor blocks into a new block, and the new block's size is the sum of two old blocks'.
- split one block into two new blocks, and you can assign the size of each block, but the sum should be equal to the old block.

Now the CEO wants to know the minimum operations to re-arrange current blocks into KKblock with equal size, please help him.

InputFirst line contains an integer TT, which indicates the number of test cases.

Every test case begins with one line which two integers NN and KK, which is the number of old blocks and new blocks.

The second line contains NN numbers a1a1, a2a2, ⋯⋯, aNaN, indicating the size of current blocks.

Limits
1≤T≤1001≤T≤100
1≤N≤1051≤N≤105
1≤K≤1051≤K≤105
1≤ai≤1051≤ai≤105 OutputFor every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the minimum operations.

If the CEO can't re-arrange KK new blocks with equal size, y equals -1. Sample Input

Sample Output

Case #1: -1
Case #2: 2
Case #3: 3

123

【体会】

第一道题，这道题，感觉自己真的要被逼疯了，组队赛时，我敲的一直WA 直到现在我也没有发现 WA 在哪里，思路完全是正确的；

赛后，我从百度拿到正确代码，用Shell 脚本随机生成 1000多组数据测试用我的代码同正确代码比较

mmp 一模一样，我不想说什么了

发一下自己 WA的代码如果有人看到，看出BUG 请赐教！！

【思路】

只有两个方式合并拆分，并且是相邻的

那么就从第一项开始，如果比平均大则减若小则加后面的项，相同不管；

我的思路是用 k的数量来控制

给 n，k 如果可以，那么一定能够形成新的 k个数因此用k 来控制

【正确代码】

//#include <bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%d\n",n)
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r

using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MAXN=100005;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};

ll inv[maxn*2];
inline void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);};}
inline ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}
inline ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}
inline ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}
inline ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}
inline ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}
inline ll inv2(ll b){return qpow(b,MOD-2);}

ll a[maxn];
int main()
{
    int t;
    int n,k;
    int cont=0;
    cin>>t;
    while(t--)
    {
        mem(a,0);
        scanf("%d %d",&n,&k);
        ll sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            sum+=a[i];
        }
        printf("Case #%d: ",++cont);
        if(sum%k!=0)
        {
            printf("-1\n");
            continue;
        }
        ll t=sum/k;
        ll last=a[1];
        int i=1;
        ll ans=0;
        ll cot=0;
        while(cot<k)
        {
            if(last>t)
            {
                last=last-t;
                ans++;
                cot++;
            }
            else if(last<t)
            {
				last +=a[++i];
				ans++;
            }
            else 
            {
                last=a[++i];
                cot++;
            }
        }
       // printf("%d %d  || Case #%d: %I64d\n",n,k,++cont,ans);
        printf("%lld\n",ans);
    }
    return 0;
}

【WA 代码请赐教】

//#include <bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%d\n",n)
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r

using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MAXN=100005;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};

ll inv[maxn*2];
inline void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);};}
inline ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}
inline ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}
inline ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}
inline ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}
inline ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}
inline ll inv2(ll b){return qpow(b,MOD-2);}

ll a[maxn];
int main()
{
    int t;
    int n,k;
    int cont=0;
    cin>>t;
    while(t--)
    {
        mem(a,0);
        scanf("%d %d",&n,&k);
        ll sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            sum+=a[i];
        }
        printf("Case #%d: ",++cont);
        if(sum%k!=0)
        {
            printf("-1\n");
            continue;
        }
        ll t=sum/k;
        ll last=a[1];
        int i=1;
        ll ans=0;
        int cot=0;
        while(cot<k)
        {
            if(last>t)
            {
                int sum=last;
                while(sum>t)
                {
                    sum-=t;
                    cot++;
                    ans++;
                }
                last=sum;
            }
            else if(last<t)
            {
                int sum=last;

                if(sum+a[i+1]>t)
                {
                    last= a[++i]-(t-sum);
                    ans+=2;
                    cot++;
                }
                else
                {
                    sum+=a[++i];
                    ans++;
                    last=sum;
                }
            }
            else if(last==t)
            {
                last=a[++i];
                cot++;
            }
        }
       // printf("%d %d  || Case #%d: %I64d\n",n,k,++cont,ans);
        printf("%lld\n",ans);
    }
    return 0;
}

posted @ 2017-10-21 21:57 Sizaif 阅读(141) 评论(0) 编辑收藏举报

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2016 CCPC Hangzhou Onsite A - ArcSoft's Office Rearrangement(HDU 5933) (贪心，模拟）

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