#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <vector>
#include <limits.h> // INT_MIN
using namespace std;
#define MAX_NUM 101
int solution(int n, int weight[]) {
    int sum = 0, h_sum, h_n = (n + 1) / 2;
    for (int i = 0; i < n; i++) sum += weight[i];
    h_sum = sum / 2;
    vector<vector<int>> dp(h_sum + 1, vector<int>(h_n + 1, INT_MIN));
    for (int i = 0; i <= h_sum; i++) dp[i][0] = 0;
    for (int i = 0; i < n; i++) {
        auto tmp = dp; // 避免覆盖！
        for (int j = weight[i]; j <= h_sum; j++) {
            for (int k = 1; k <= h_n; k++)
                tmp[j][k] = max(tmp[j][k], dp[j-weight[i]][k-1]+weight[i]);
        }
        dp = tmp;
    }
    if (n % 2) return sum - 2 * max(dp[h_sum][h_n], dp[h_sum][h_n-1]);
    else return sum - 2 * dp[h_sum][h_n];
}

int main()
{    
	string str("");
	getline(cin, str);
	int a[MAX_NUM];
	int i = 0;
	char *p;
	int count = 0;
	const char* strs = str.c_str();
	p = strtok((char *)strs, " ");
	while(p) {
		a[i] = atoi(p);
		count++;
		p = strtok(NULL, " ");
		i++;
		if(i >= MAX_NUM)
			break;
	}
	int result = solution(count, a);
	cout << result << endl;
	return 0;
}

报数

• 将N（N<10000）个人排成一排，从第1个人开始报数；如果报数是M的倍数就出列，报到队尾后则回到队头继续报，直到所有人都出列；
• 按照出列顺序为每个人依次分配工号

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
#include <vector>
using namespace std;
typedef struct _node
{
    int num;
    struct _node * next;
}node;
void solution(int N, int M)
{
	if (N < 1 || M < 1) return;
    vector<int> rows(N, 0);
    for (int i = 0; i < N; i++) rows[i] = i + 1;
    int pos = 0;
    while (rows.size() > 0) {
        pos += M - 1;
        pos %= rows.size();
        cout << rows[pos] << " ";
        rows.erase(rows.begin() + pos);
    } 
    cout << endl;
}
int main()
{
	int N;
	int M;
	string str("");
	getline(cin, str);
	char *p;
	const char* strs = str.c_str();
	p = strtok((char *)strs, " ");
	N = atoi(p);
	p = strtok(NULL, " ");
	M = atoi(p);
	solution(N, M);
	return 0;
}

#include <iostream>
#include <stdlib.h>
#include <string.h>
using namespace std;
int solution(int a[], int N)
{
    int pre = 0, cur = 0;
    int steps = 0;
    for (int i = 0; i < N - 1; i++) {
        cur = max(cur, i+a[i]);
        if (pre == i) {
            pre = cur;
            steps++;
            if (cur >= N - 1) 
                break;
        }
    }
    return pre >= N - 1 ? steps : -1;
}
int main()
{
	string str("");
	getline(cin, str);
	int a[2000];
	int i = 0;
	char *p;
	int count = 0;
	const char* strs = str.c_str();
	p = strtok((char *)strs, " ");
	while(p)
	{
		a[i] = atoi(p);
		count++;
		p = strtok(NULL, " ");
		i++;
		if(i >= 2000)
			break;
	}
	int num = solution(a, count);
	cout << num << endl;
	return 0;
}

//write code here
vector<vector<vector<int>>> dp(countOfApp + 1,vector<vector<int