hdu 4811 数学 不难

http://acm.hdu.edu.cn/showproblem.php?

pid=4811

由于看到ball[0]>=2 && ball[1]>=2 && ball[2]>=2  ans=(sum-6)*6+15    sum是三种颜色的球个数的和，然后就想到分类讨论，由于情况是可枚举的。

发现整数假设不加LL直接用%I64d打印会出问题

//#pragma comment(linker, "/STACK:102400000,102400000")

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <string>

#include <iostream>

#include <iomanip>

#include <cmath>

#include <map>

#include <set>

#include <queue>

using namespace std;

#define ls(rt) rt*2

#define rs(rt) rt*2+1

#define ll long long

#define ull unsigned long long

#define rep(i,s,e) for(int i=s;i<e;i++)

#define repe(i,s,e) for(int i=s;i<=e;i++)

#define CL(a,b) memset(a,b,sizeof(a))

#define IN(s) freopen(s,"r",stdin)

#define OUT(s) freopen(s,"w",stdout)

const ll ll_INF = ((ull)(-1))>>1;

const double EPS = 1e-8;

const double pi = acos(-1.0);

const int INF = 100000000;

ll ball[3];

int main()

{

    while(~scanf("%I64d%I64d%I64d",&ball[0],&ball[1],&ball[2]))

    {

        sort(ball, ball+3);

        ll sum=ball[0]+ball[1]+ball[2];

        if(ball[0]>=2 && ball[1]>=2 && ball[2]>=2)

        {

            printf("%I64d\n",(ll)(sum-6)*6+15);

            continue;

        }

        if(!ball[0] && !ball[1])

        {

            printf("%I64d\n",ball[2]==0||ball[2]==1 ? 0LL :2*ball[2]-3);

            continue;

        }

        if(!ball[0])

        {

            ll ans=0LL;

            if(ball[1]==1){printf("%I64d\n",(ans=ball[2]==1?1:(ll)(3*ball[2])-3));continue;}

            if(ball[1]>=2){printf("%I64d\n",(ans=4*(sum-4)+6));continue;}

        }

        if(ball[0] == 1)

        {

            if(ball[1]==1)//printf("%I64d\n", ball[2]==1 ? 3LL : 3*ball[2]);

            {

                if(ball[2] == 1)puts("3");

                if(ball[2] >=2) printf("%I64d\n",6+4*(ball[2]-2));

            }

            if(ball[1]>1)printf("%I64d\n", (sum-5)*5+10);

        }

    }

    return 0;

}

AC了之后看到别人的代码真是短啊......

http://blog.csdn.net/accelerator_/article/details/25918093

推理一下，发现能够先求出后面放小球能够加分的最大值，然后前面的和为0 + 1 + 2 + ...+ max,max最大为6，由于每一个球最多算左右两边

代码:

#include <iostream>

#include <algorithm>

using namespace std;

long long a, b, c;

long long tra(long long num) {

    return num > 2 ? 2 : num;

}

int main() {

    while (cin >> a >> b >> c) {

    long long sum = tra(a) + tra(b) + tra(c);

    long long hav = max(0ll, a + b + c - sum);

    cout << (hav * sum + (sum - 1) * sum / 2) << endl;

    }

    return 0;

}