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;
}