250 题：
在一个矩阵中取出一个点，求不包含那个点的权值最大的矩阵。
直接枚举四种情况。
/*
收获：
*/
#include<iostream>
#include<cstdlib>
#include<vector>
#include<map>
#include<cstring>
#include<set>
#include<string>
#include<algorithm>
#include<sstream>
#include<ctype.h>
#include<fstream>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<stack>
#include<queue>
#include<ctime>
//#include<conio.h>
using namespace std;
const int INF_MAX=0x7FFFFFFF;
const int INF_MIN=-(1<<30);
const double eps=1e-10;
const double pi=acos(-1.0);
#define pb push_back //a.pb( )
#define chmin(a,b) ((a)<(b)?(a):(b))
#define chmax(a,b) ((a)>(b)?(a):(b))
template<class T> inline T gcd(T a,T b)//NOTES:gcd(
{if(a<0)return gcd(-a,b);if(b<0)return gcd(a,-b);return (b==0)?a:gcd(b,a%b);}
template<class T> inline T lcm(T a,T b)//NOTES:lcm(
{if(a<0)return lcm(-a,b);if(b<0)return lcm(a,-b);return a*(b/gcd(a,b));}
typedef pair<int, int> PII;
typedef vector<PII> VPII;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef long long LL;
int dir_4[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
int dir_8[8][2]={{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1},{1,0},{1,1}};
//下，左下，左，左上，上，右上，右，右下。
//******* WATER ****************************************************************
//flag = 0，横；flag = 1,竖
int calc(vector <string> flowers, bool flag, int start, int end) {
int ret = 0;
if(start > end) return 0;
if(!flag) {
for(int j = 0; j < flowers[0].length(); j++) {
for(int i = start; i <= end; i++) {
if(flowers[i][j] == 'F') ret++;
}
}
}
else {
for(int i = 0; i < flowers.size(); i++) {
for(int j = start; j <= end; j++) {
if(flowers[i][j] == 'F') ret++;
}
}
}
return ret;
}
class FoxAndFlowerShopDivTwo {
public:
int theMaxFlowers(vector <string> flowers, int r, int c) {
int ret = 0, tmp;
tmp = calc(flowers, false, 0, r - 1);
ret = chmax(tmp, ret);
tmp = calc(flowers, false, r + 1, flowers.size() - 1);
ret = chmax(tmp, ret);
tmp = calc(flowers, true, 0, c - 1);
ret = chmax(tmp, ret);
tmp = calc(flowers, true, c + 1, flowers[0].length() - 1);
ret = chmax(tmp, ret);
return ret;
}
};
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500:
题目要求对一个三角形的一堆球用三种颜色染色，相邻的球的颜色不同，注意会有规律：。
1
2 3
3 1 2
1 2 3 1
2 3 1 2 3
3 1 2 3 1 2
球的个数有两种情况：
1.所有的1,2,3中颜色的球的个数相同，这时候N%3 == 2；
2.所有的1,2,3中颜色1的多一个，颜色2和3的相同。颜色1的个数为ceil(sum/3),颜色2和3的个数为floor（sum/3)，其中sum = （N + 1) * N ;
RGB三种颜色最多各染r，g，b种，那么有两种情况：
1. r + g + b总数不够染色，限制染色数目，为:floor((r +g+b)/sum);
2.最少的颜色限制染色数目，floor(min_color / ceil(sum/3));其中min_color = min(r, g, b);
code：
/*
收获：
*/
#include<iostream>
#include<cstdlib>
#include<vector>
#include<map>
#include<cstring>
#include<set>
#include<string>
#include<algorithm>
#include<sstream>
#include<ctype.h>
#include<fstream>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<stack>
#include<queue>
#include<ctime>
//#include<conio.h>
using namespace std;
const int INF_MAX=0x7FFFFFFF;
const int INF_MIN=-(1<<30);
const double eps=1e-10;
const double pi=acos(-1.0);
#define pb push_back //a.pb( )
#define chmin(a,b) ((a)<(b)?(a):(b))
#define chmax(a,b) ((a)>(b)?(a):(b))
template<class T> inline T gcd(T a,T b)//NOTES:gcd(
{if(a<0)return gcd(-a,b);if(b<0)return gcd(a,-b);return (b==0)?a:gcd(b,a%b);}
template<class T> inline T lcm(T a,T b)//NOTES:lcm(
{if(a<0)return lcm(-a,b);if(b<0)return lcm(a,-b);return a*(b/gcd(a,b));}
typedef pair<int, int> PII;
typedef vector<PII> VPII;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef long long LL;
int dir_4[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
int dir_8[8][2]={{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1},{1,0},{1,1}};
//下，左下，左，左上，上，右上，右，右下。
//******* WATER ****************************************************************
class FoxPaintingBalls {
public:
long long theMax(long long R, long long G, long long B, int N) {
LL minx = chmin(R, chmin(G, B));
LL tot = (N + 1) * N / 2;
LL one = tot / 3;
if(N == 1) return R+G+B;
return chmin((R+G+B) / tot, minx / one);
}
};
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1000分，目前不会，等待题解中。。。