第一题：小明种苹果
n , m = map(int , input().split())
t , k , p = 0 , 0 , -1
for _ in range(n):
l = list(map(int , input().split()))
t += sum(l)
x = -sum(l[i] for i in range(1 , len(l)))
if x > p:
p = x
k = _ + 1
print(t , k , p)
第二题：小明种苹果（续）
#include
#include
#include
using namespace std;
typedef long long ll;
const int N = 1010;
int n;
ll t;
int d , e;
bool diao[N];
int main()
{
memset(diao , 0 , sizeof diao);
cin >> n;
for(int i = 0;i < n;i ++)
{
int m;
cin >> m;
vectorv;
for(int j = 0;j < m;j ++)
{
ll x;
cin >> x;
v.push_back(x);
}
ll temp = v[0];
// 前缀和的思想
for(int j = 1;j < m;j ++)
{
if(v[j] <= 0) temp += v[j];
else
{
if(temp != v[j]) diao[i] = true;
}
}
for(int j = m - 1;j >= 0;j --)
{
t += v[j];
if(v[j] > 0) break;
}
}
for(int i = 0;i < n;i ++)
if(diao[i]) d ++;
for(int i = 0;i < n;i ++)
if(diao[i] && diao[(i + 1) % n] && diao[(i + 2) % n]) e ++;
cout << t << " " << d << " " << e << endl;
}
第三题：字符画
#include
#include
#include
using namespace std;
typedef unsigned char UC;
const int N = 1080, M = 1920;
int m, n, p, q;
UC g[N][M][3];
inline int get(char c)
{
if (c <= '9') return c - '0';
return c - 'a' + 10;
}
inline char get(int x)
{
if (x <= 9) return x + '0';
return x - 10 + 'A';
}
inline void print(char* str)
{
for (int i = 0; str[i]; i ++ )
printf("\\x%c%c", get(str[i] / 16), get(str[i] % 16));
}
int main()
{
scanf("%d%d%d%d", &m, &n, &p, &q);
char str[100];
for (int i = 0; i < n; i ++ )
for (int j = 0; j < m; j ++ )
{
scanf("%s", str);
int len = strlen(str);
if (len == 2)
{
int t = get(str[1]);
for (int k = 0; k < 3; k ++ )
g[i][j][k] = t * 16 + t;
}
else if (len == 4)
{
for (int k = 0; k < 3; k ++ )
{
int t = get(str[1 + k]);
g[i][j][k] = t * 16 + t;
}
}
else
{
for (int k = 0; k < 3; k ++ )
g[i][j][k] = get(str[1 + k * 2]) * 16 + get(str[2 + k * 2]);
}
}
int bg[3] = {0};
for (int i = 0; i < n / q; i ++ )
{
for (int j = 0; j < m / p; j ++ )
{
int cur[3] = {0};
for (int x = 0; x < q; x ++ )
for (int y = 0; y < p; y ++ )
for (int z = 0; z < 3; z ++ )
cur[z] += g[i * q + x][j * p + y][z];
for (int k = 0; k < 3; k ++ ) cur[k] /= p * q;
if (cur[0] == bg[0] && cur[1] == bg[1] && cur[2] == bg[2]) ; // pass
else if (!cur[0] && !cur[1] && !cur[2]) print("\033[0m");
else
{
sprintf(str, "\033[48;2;%d;%d;%dm", cur[0], cur[1], cur[2]);
print(str);
}
for (int k = 0; k < 3; k ++ ) bg[k] = cur[k];
print(" ");
}
if (bg[0] || bg[1] || bg[2])
{
print("\033[0m");
for (int k = 0; k < 3; k ++ ) bg[k] = 0;
}
print("\n");
}
return 0;
}
第四题：推荐系统
#include
#include
#include
#include
#include
#include
第五题：城市规划
#include
#include
#include
using namespace std;
typedef long long LL;
const int N = 50010, M = N * 2;
int n, m, K;
int h[N], e[M], w[M], ne[M], idx;
LL f[N][110];
bool st[N];
int sz[N];
LL ans = 1e18;
void add(int a, int b, int c)
{
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}
void dfs(int u, int fa)
{
f[u][0] = 0;
if (st[u]) f[u][1] = 0;
sz[u] = 1;
for (int i = h[u]; ~i; i = ne[i]) // 枚举物品组
{
int ver = e[i];
if (ver == fa) continue;
dfs(ver, u);
sz[u] += sz[ver];
for (int j = min(sz[u], K); j >= 0; j -- ) // 枚举体积
for (int k = 0; k <= min(j, sz[ver]); k ++ ) // 枚举决策
f[u][j] = min(f[u][j], f[u][j - k] + f[ver][k] + (LL)w[i] * k * (K - k));
}
ans = min(ans, f[u][K]);
}
int main()
{
scanf("%d%d%d", &n, &m, &K);
memset(h, -1, sizeof h);
while (m -- )
{
int x;
scanf("%d", &x);
st[x] = true;
}
for (int i = 0; i < n - 1; i ++ )
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(a, b, c), add(b, a, c);
}
memset(f, 0x3f, sizeof f);
dfs(1, -1);
printf("%lld\n", ans);
return 0;
}
