# T1

题目大意

提交代码

分类讨论

可以枚举三个数中最小

下面以 $a$ 最小的情况为例

  • $$\because 7a+4b+3c=n$$
  • $$\therefore 7a=n-4b-3c$$
  • 由于我们要考虑 $a+b+c$ 最大. 因为 $b$ 的系数比 $c$ 要大, 而和为定值, 所以当 $a$ 固定时, $b$ 越小越好
  • 所以我们就可以令 $b$ 为 $a, a+1, a+2$ 中的一个(为了满足 $c$ 是整数, 找一个值使得 $n-4b-7a \equiv 0 \pmod {3}$)

将上面的方法重复套到 $b$ 和 $c$ 中, 跑三遍 for (有点丑)即可

最后然后别忘了校验得出来的值是否合理

完整代码
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#include <cstdio>
#include <iostream>

#define min3(a, b, c) std::min(std::min(a, b), c)

const int coeff_a = 7, coeff_b = 4, coeff_c = 3;

int main()
{
int n;
int ans_a = -1, ans_b = -1, ans_c = -1;

scanf("%d", &n);

for (int a = 0; a <= n; a++) // a <= b, c
{
int b = a, c;

while (((n - a * coeff_a - b * coeff_b) % coeff_c + coeff_c) % coeff_c != 0)
b++;
c = (n - a * coeff_a - b * coeff_b) / coeff_c;

if (c < 0 || c < a)
break; // 矛盾

if (min3(ans_a, ans_b, ans_c) < a)
ans_a = a, ans_b = b, ans_c = c;
else if (min3(ans_a, ans_b, ans_c) == a && ans_a + ans_b + ans_c < a + b + c)
ans_a = a, ans_b = b, ans_c = c;
}

for (int b = 0; b <= n; b++) // b <= a, c
{
int a = b, c;

while (((n - a * coeff_a - b * coeff_b) % coeff_c + coeff_c) % coeff_c != 0)
a++;
c = (n - a * coeff_a - b * coeff_b) / coeff_c;

if (c < 0 || c < b)
break; // 矛盾

if (min3(ans_a, ans_b, ans_c) < b)
ans_a = a, ans_b = b, ans_c = c;
else if (min3(ans_a, ans_b, ans_c) == b && ans_a + ans_b + ans_c < a + b + c)
ans_a = a, ans_b = b, ans_c = c;
}

for (int c = 0; c <= n; c++) // c <= a, b
{
int a = c, b;

while (((n - a * coeff_a - c * coeff_c) % coeff_b + coeff_b) % coeff_b != 0)
a++;
b = (n - a * coeff_a - c * coeff_c) / coeff_b;

if (b < 0 || b < c)
break; // 矛盾

if (min3(ans_a, ans_b, ans_c) < c)
ans_a = a, ans_b = b, ans_c = c;
else if (min3(ans_a, ans_b, ans_c) == c && ans_a + ans_b + ans_c < a + b + c)
ans_a = a, ans_b = b, ans_c = c;
}

if (ans_a == -1)
puts("-1");
else
printf("%d %d %d", ans_a, ans_b, ans_c);
return 0;
}

# T2 & T3

留坑待填


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