SRM401div2题解

DIV2：

250：水题，题意是给定两个点的坐标，问连接这俩个点坐标的线段经过了多少整点 解法：gcd(x2-x1, y2-y1)-1

500：dp，题意是给定一个值fieldOrder，求非增序列且第k个数小于fieldOrder-k+1的个数  解法：dp[i][j] 表示第i个数大小是j的时候序列的个数 dp[i+1][k] += dp[i][j](k<=j,j<=fieldOrder-i+1)结果就是dp[i][j](j<=i)所有的和

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

long long dp[40][40];

class FIELDDiagrams {
    public:
        long long countDiagrams(int fieldOrder) {
            long long ans = 0;
            for (int i = 1; i <= fieldOrder; i++)
                dp[1][i] = 1;
            for (int i = 1; i <= fieldOrder; i++)
                for (int j = 1; j <= fieldOrder-i+1; j++)
                    for (int k = 1; k <= j; k++)
                        dp[i+1][k] += dp[i][j];
            for (int i = 1; i <= fieldOrder; i++) 
                for (int j = 1; j <= fieldOrder-i+1; j++)
                    ans += dp[i][j];
            return ans;
        }
};

1000：