1001 数组中和等于K的数对 1090 3个数和为0

二分查找。对数组每个V[i],在其中查找K-V[i],查找完成后修改v[i]避免重复输出

#include<iostream>

#include<algorithm>

#include<vector>

#include<cstdio>

using namespace std;

typedef long long LL;

#define INF 0x3f3f3f3f

int main()

{

    vector<LL> vv;

    bool sign = false;

    LL t,k,n;

    scanf("%lld%lld",&k,&n);

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

    {

        scanf("%lld",&t);

        vv.push_back(t);

    }

    sort(vv.begin(),vv.end());

    int l =vv.size();

    for(int i=;i<l;i++)

    {

        if(vv[i]==INF) continue;

        vector<LL>::iterator it;

        if(binary_search(vv.begin(),vv.end(),k-vv[i]))

            it = lower_bound(vv.begin(),vv.end(),k-vv[i]);

        else

            continue;

        if(it!=vv.end()&&*it!=vv[i])

        {

            sign = true;

            if(*it>vv[i])

                printf("%lld %lld\n",vv[i],*it);

            else

                printf("%lld %lld\n",*it,vv[i]);

            *it = INF;

        }

    }

    if(!sign)

        printf("No Solution\n");

    return ;

}

三个数的和为0，在前面代码基础上略作修改即可。

#include<iostream>

#include<algorithm>

#include<vector>

#include<cstring>

#include<cstdio>

using namespace std;

typedef long long LL;

#define MAXN 1001

#define INF 0x3f3f3f3f

bool been[MAXN];

bool all_sign = false;

LL n;

vector<LL> VV;

void solve(LL k,vector<LL> vv,LL pos)

{

    memset(been,false,sizeof(been));

    for(LL i=pos+;i<n;i++)

    {

        if(been[i]==true) continue;

        vector<LL>::iterator it;

        if(binary_search(vv.begin()+pos+,vv.end(),k-vv[i]))

            it = lower_bound(vv.begin()+pos+,vv.end(),k-vv[i]);

        else

            continue;

        if(it!=vv.end()&&*it!=vv[i])

        {

            all_sign = true;

            if(*it>vv[i])

                printf("%lld %lld %lld\n",-k,vv[i],*it);

            else

                printf("%lld %lld %lld\n",-k,*it,vv[i]);

            been[it-vv.begin()] = true;

        }

    }

}

int main()

{

    LL temp,i;

    scanf("%lld",&n);

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

    {

        scanf("%lld",&temp);

        VV.push_back(temp);

    }

    sort(VV.begin(),VV.end());

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

    {

        if(VV[i]>=)

            break;

        temp = VV[i];

        solve(-temp,VV,i);

        been[i] = true;

    }

    if(!all_sign)

        printf("No Solution\n");

    return ;

}