- 伯劳的问题
|a|，|b|都小于 10^13，超出 int（10^9）范围，所以使用 long long (10^18)
int main()
{
int n;
cin >> n;
while (n–)
{
ll a, b;
cin >> a >> b;
cout << a + b << endl;
}
return 0;
} - 宝可梦大师
数组循环求和；输入时直接判断奇偶累加
int main()
{
int a = 0, b = 0;
int n;
scanf("%d", &n);
int arr[maxn];
for(int i = 0; i < n; ++i)
{
scanf("%d", &arr[i]);
if(arr[i] % 2 == 0) a += arr[i];
else b += arr[i];
}
cout << a << ’ ’ << b;
return 0;
} - 联合作战计划
由题意易得，若序列无相邻元素即输出 1，否则输出 2
int main()
{
int q;
scanf("%d", &q);
while(q–)
{
int n;
int arr[maxn];
scanf("%d", &n);
for(int i = 0; i < n; ++i)
scanf("%d", &arr[i]);
if(n == 1)
{
printf(“1\n”);
continue;
}
sort(arr, arr + n);
int flag = 0;
for(int i = 1; i < n; ++i)
{
if(arr[i] - arr[i - 1] == 1)
{
printf(“2\n”);
flag = 1;
break;
}
}
if(!flag)
printf(“1\n”);
}
return 0;
} - 镜子大师
先将三个数排序，将问题转换成如何把最小值分配给最大值与次大值
易得如果次大值与最小值相加大于等于最大值则答案为三者求和除 2，否则答案为最小
值与次大值的和
bool cmp(int x, int y)
{
return x > y;
}
int main()
{
int T;
scanf("%d", &T);
while(T–)
{
int a[3];
scanf("%d %d %d", &a[0], &a[1], &a[2]);
sort(a, a + 3, cmp);
int ans = 0;
if(a[1] + a[2] >= a[0]) ans = (a[0] + a[1] + a[2]) / 2;
else ans = min(a[0], a[1] + a[2]);
cout << ans << endl;
}
} - 绝地武士
由题意易得，我们只需要得出能力值相同中人数的最大值
数据范围 q≤10^30，显然需要使用字符串，注意前置 0
int main()
{
int n;
while(scanf("%d", &n) != EOF)
{
map<string, int> mp;
for(int i = 0; i < n; ++i)
{
string ss;
cin >> ss;
for(string :: iterator its = ss.begin(); *its == ‘0’; ss.erase(its));
mp[ss]++;
}
map<string, int> :: iterator it;
int sum = 0;
for(it = mp.begin(); it != mp.end(); ++it)
{
sum = max(sum, it->second);
}
printf("%d\n", sum);
}
}
6. 织梦岛
BFS 模板题，二维变三维
int L, R, C;
struct Maze
{
int x, y, z;
}m;
char maze[35][35][35];
int d[35][35][35];
int sx, sy, sz, ex, ey, ez;
int dx[6] = {1, -1, 0, 0, 0, 0}, dy[6] = {0, 0, 1, -1, 0, 0}, dz[6] = {0, 0, 0, 0, 1, -1};
int bfs()
{
queue q;
for (int i = 0; i < L; ++i)
{
for (int j = 0; j < R; ++j)
{
for (int k = 0; k < C; ++k)
d[i][j][k] = INF;
}
}
d[sx][sy][sz] = 0;
m.x = sx, m.y = sy, m.z = sz;
q.push(m);
while (q.size())
{
Maze temp = q.front();
q.pop();
if (temp.x == ex && temp.y == ey && temp.z == ez) break;
for (int i = 0; i < 6; ++i)
{
int nx = temp.x + dx[i], ny = temp.y + dy[i], nz = temp.z + dz[i];
if ((nx >= 0 && nx < L) && (ny >= 0 && ny < R) && (nz >= 0 && nz <
C) && d[nx][ny][nz] == INF && maze[nx][ny][nz] != ‘#’)
{
Maze ans;
ans.x = nx, ans.y = ny, ans.z = nz;
q.push(ans);
d[nx][ny][nz] = d[temp.x][temp.y][temp.z] + 1;
}
}
}
return d[ex][ey][ez];
}
int main()
{
while (cin >> L >> R >> C)
{
if (L == 0 && R == 0 && C == 0) break;
for (int i = 0; i < L; ++i)
{
for (int j = 0; j < R; ++j)
{
for (int k = 0; k < C; ++k)
{
cin >> maze[i][j][k];
if (maze[i][j][k] == ‘S’)
sx = i, sy = j, sz = k;
if (maze[i][j][k] == ‘E’)
ex = i, ey = j, ez = k;
}
}
getchar();
}
int num = bfs();
if (num != INF)
cout << num << endl;
else
cout << “YOU DIED” << endl;
}
return 0;
}
7. 谍战风云
由题意可得，将数转换为三进制，系数需为 0 或 1，若三进制数中存在 2，则不满
足条件
寻找三进制数中最高位的 2，将低位全置为 0 即解
ll a1[105], a2[105];
ll func(ll n)
{
memset(a1, 0, sizeof(a1));
memset(a2, 0, sizeof(a2));
int flag = 0;
while(n)
{
a1[flag++] = n % 3;
n /= 3;
}
flag++;
a2[0] = 0;
for(int i = 1; i < flag; ++i)
{
a2[i] = a1[flag - i - 1];
}
int p = 0;
for(int i = 1; i < flag; ++i)
{
if(a2[i] == 2)
{
p = i;
break;
}
}
for(int i = p - 1; i >= 0; --i)
{
if(a2[i] == 0)
{
a2[i] = 1;
for(int j = i + 1; j < flag; ++j)
a2[j] = 0;
break;
}
}
ll num = a2[0];
for(int i = 0; i < flag - 1; ++i)
num = num * 3 + a2[i + 1];
return num;
}
int main()
{
int n;
scanf("%d", &n);
while(n–)
{
ll t;
scanf("%lld", &t);
ll ans = func(t);
printf("%lld\n", ans);
}
}
