标签:Multi Chinese Contest int res cnt ++ mp mod
A:Character Encoding
插板法+容斥
https://blog.csdn.net/codeswarrior/article/details/81906367?utm_medium=distribute.pc_relevant_t0.none-task-blog-2~default~BlogCommendFromBaidu~default-1.control&depth_1-utm_source=distribute.pc_relevant_t0.none-task-blog-2~default~BlogCommendFromBaidu~default-1.control
讲的特别好
int fact[maxn], finv[maxn], inv[maxn];
void init() {
fact[0] = fact[1] = finv[0] = finv[1] = inv[0] = inv[1] = 1;
for(int i = 2; i < maxn; ++ i) {
fact[i] = fact[i - 1] * i % mod;
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
finv[i] = finv[i - 1] * inv[i] % mod;
}
}
inline int C(int n, int m) {
if(n < 0 || m < 0 || m > n) return 0;
return fact[n] * finv[m] % mod * finv[n - m] % mod;
}
inline void qmod(int &a) {
while(a >= mod) a -= mod;
}
signed main() {
init();
int t = rd();
while(t--) {
int n = rd(), m = rd(), k = rd();
int ans = C(k + m - 1, m - 1);
for(int i = 1, f = -1; i * n <= k; ++ i, f = -f)
qmod(ans += f * C(k - i * n + m - 1, m - 1) * C(m, i) % mod + mod);
printf("%lld\n", ans);
}
return 0;
}
D: Parentheses Matrix
打标找规律的构造题,规模比较大,注意的是剪枝
int ans = 0, n, m, cnt = 0;
int c[maxn], r[maxn];
char mp[202][202];
void dfs(int x, int y) {
cnt ++;
if(cnt % 100000000 == 0) printf("%lld\n", cnt);
if(x == n - 1) {
int res = 0;
for(int i = 0; i < n; ++ i) if(r[i]) res ++;
for(int i = 0; i < m; ++ i) if(c[i]) res ++;
if(n + m - res < ans) return ;
res = 0;
for(int i = 0; i < n; ++ i) {
int cnt = 0;
for(int j = 0; j < m; ++ j) if(mp[i][j] == '(') cnt ++;
else if(cnt) cnt --;
else {cnt = -1; break;}
if(cnt == 0) res ++;
}
for(int i = 0; i < m; ++ i) {
int cnt = 0;
for(int j = 0; j < n; ++ j) if(mp[j][i] == '(') cnt ++;
else if(cnt) cnt --;
else {cnt = -1; break;}
if(cnt == 0) res ++;
}
//if(res == ans) {
// for(int i = 0; i < n; ++ i) puts(mp[i]);
// dbg(ans); puts("");
//}
if(res > ans) {
for(int i = 0; i < n; ++ i) {
for(int j = 0; j < m; ++j) printf("%c", mp[i][j]);
puts("");
}
puts("");
ans = res;
dbg(ans);
}
return;
}
mp[x][y] = '('; r[x] ++; c[y] ++;
dfs(x + (y + 1) / m, (y + 1) % m);
mp[x][y] = '('; r[x] --; c[y] --;
mp[x][y] = ')'; r[x] --; c[y] --;
dfs(x + (y + 1) / m, (y + 1) % m);
mp[x][y] = ')'; r[x] ++; c[y] ++;
}
signed main() {
int t = rd();
while(t--) {
n = rd(), m = rd();
for(int i = 0; i <= n; ++ i) r[i] = 0; r[0] = r[n - 1] = m;
for(int i = 0; i <= m; ++ i) c[i] = 0;
for(int i = 0; i <= n; ++ i) mp[i][m] = '\0';
//ans = 0;
ans = 8;
for(int j = 0; j < m; ++ j) mp[0][j] = '(', mp[n - 1][j] = ')';
//dfs(1, 0);
//dbg(ans);
if(n & m & 1) for(int i = 1; i <= n; ++ i) for(int j = 0; j <= m; ++ j) printf("%c", "(\n"[j == m]);
else if(n & 1) for(int i = 1; i <= n; ++ i) {for(int j = 0; j < m; ++ j) printf("%c", "()"[j % 2]); puts("");}
else if(m & 1) for(int i = 1; i <= n; ++ i) {for(int j = 0; j < m; ++ j) printf("%c", "()"[i % 2 == 0]); puts("");}
else {
int x = n, y = m;
if(x > y) swap(x, y);
for(int j = 0; j < y; ++ j) mp[0][j] = '(', mp[x - 1][j] = ')';
if(x == 4) for(int j = 0; j < y; ++ j) mp[1][j] = "()"[j >= (y >> 1)], mp[2][j] = ")("[j >= (y >> 1)];
else if(x > 4){
for(int i = 1; i < x - 1; ++ i) {
if(i & 1) for(int j = 0; j < y; ++ j) mp[i][j] = "()"[j % 2];
else for(int j = 0; j < y; ++ j) mp[i][j] = "()"[(j % 2 == 0 && j != 0) || j == y - 1];
}
}
if(n > m) for(int i = 0; i < x; ++ i) for(int j = i + 1; j < y; ++ j) swap(mp[i][j], mp[j][i]);
for(int i = 0; i < n; ++ i) mp[i][m] = '\0';
for(int i = 0; i < n; ++ i) puts(mp[i]);
}
}
return 0;
}
/*
case: h = 2
(((((
)))))
case: h = 4
((((((
((()))
)))(((
))))))
case: h > 6
((((((
()()()
(()())
()()()
(()())
))))))
one case can swap to another case
*/
标签:Multi,Chinese,Contest,int,res,cnt,++,mp,mod 来源: https://www.cnblogs.com/ddsszdnt/p/15021852.html
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