A Help Me！！！
#include <bits/stdc++.h>
using namespace std;
int main() {
char a[105], b;
scanf("%[^\n]", a);
int len = strlen(a);
for(int i = 0; i < len; ++i) {
if(a[i] == ' ') b = ' ';
else if(i + 2 < len && a[i + 2] == '#')
b = (a[i] - '0') * 10 + a[i + 1] - '0' + 'a' - 1, i += 2;
else
b = a[i] - '0' + 'a' - 1;
putchar(b);
}
return 0;
}
B 任意数
原题
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int mod = 1e9 + 7;
ll ksm(ll a, ll b) {
ll ans = 1;
while(b) {
if(b & 1) ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans % mod;
}
int main() {
ll n, m;
cin >> n >> m;
cout << (ksm(2, m) * (n + 1) % mod - 1) % mod;
return 0;
}
C 小L的光图
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main() {
ll n, p, k, ans = 0;
cin >> n >> p >> k;
cout << (k % n * p) % n;
return 0;
}
D 小L求偶
#include <bits/stdc++.h>
using namespace std;
int a[10005];
int main() {
int n, ans = 0;
cin >> n;
for(int i = 0; i < n; ++i) {
scanf("%d", &a[i]);
if(a[i] % 2 == 0) ans += a[i];
}
for(int i = 0; i < n; ++i) {
int val, idx;
scanf("%d%d", &val, &idx);
if(a[idx] % 2 == 0) ans -= a[idx];
a[idx] += val;
if(a[idx] % 2 == 0) ans += a[idx];
printf("%d\n", ans);
}
return 0;
}
E 小L的表达式
#include <bits/stdc++.h>
using namespace std;
int main() {
char c, pc ='(';
bool pre = 0, now = 0;
while((c = getchar()) && c != '\n') {
if(isdigit(c) || (pc == '(' && (c == '-' || c == '+')) || c == '.') {
now = 1;
putchar(c);
}
else {
now = 0;
if(pre != now) putchar('\n');
putchar(c);
putchar('\n');
}
pre = now;
pc = c;
}
return 0;
}
F 可恶的小J
#include <bits/stdc++.h>
#define dx (x + dir[i])
#define dy (y + dir[i + 1])
using namespace std;
typedef struct {
int x, y;
int t, s; // 体力，时间
} node;
int dir[] = {
0, 1, 0, -1, 0};
char a[205][205];
int vis[205][205];
int main() {
memset(vis, -1, sizeof vis);
int m, n, t;
cin >> m >> n >> t;
queue<node> q;
for(int i = 0; i < m; ++i) {
scanf("%s", a[i]);
for(int j = 0; j < n; ++j)
if(a[i][j] == '@')
q.push({
i, j, t, 0}), vis[i][j] = 0;
}
int flag = -1;
while(!q.empty()) {
auto pir = q.front();
q.pop();
int x = pir.x, y = pir.y;
int t = pir.t, s = pir.s + 1;
for(int i = 0; i < 4; ++i) {
if(dx < 0 || dy < 0 || dx >= m || dy >= n || vis[dx][dy] >= t)
continue;
if(a[dx][dy] == '+') {
flag = s;
break;
}
else if(a[dx][dy] == '#') {
vis[dx][dy] = t - 1;
q.push({
dx, dy, t - 1, s});
}
else if(a[dx][dy] == '*') {
vis[dx][dy] = t;
q.push({
dx, dy, t, s});
}
}
if(~flag) break;
}
cout << flag;
return 0;
}
G 敌对的小L和小J
#include <bits/stdc++.h>
using namespace std;
int a[1005];
bool check(int x, int y) {
if(__gcd(x, y) == 1) return 1;
return 0;
}
int main() {
int n, ans = 0;
cin >> n;
for(int i = 0; i < n; ++i) {
cin >> a[i];
for(int j = 0; j < i; ++j) {
if(check(a[i], a[j])) {
ans++;
}
}
}
cout << ans;
return 0;
}