https://codeforc.es/gym/101982/
A - Exam
很显然，要把大家做一样的优先认为大家都是对的，大家做不一样的认为自己是对的。最后假如对方有剩余的题是对的那么就要从中减去。
#include
using namespace std;
int k;
char s[1005];
char t[1005];
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
#endif // Yinku
scanf("%d%s%s", &k, s + 1, t + 1);
int n = strlen(s + 1);
int cnt = 0;
for(int i = 1; i <= n; ++i) {
if((s[i] == t[i])) {
if(k) {
++cnt;
--k;
}
} else {
++cnt;
}
}
printf("%d\n", cnt - k);
} B - Coprime Integers
求范围内的互质数对，莫比乌斯反演的逗比模板题。而且还是线性筛就够了。小心溢出。
#include
using namespace std;
typedef long long ll;
int n;
const int MAXN = 1e7;
int sumu[MAXN + 5];
int pri[MAXN + 5], pritop;
bool notpri[MAXN + 5];
void sieve(int n) {
notpri[1] = sumu[1] = 1;
for(int i = 2; i <= n; i++) {
if(!notpri[i])
pri[++pritop] = i, sumu[i] = -1;
for(int j = 1; j <= pritop && i * pri[j] <= n; j++) {
notpri[i * pri[j]] = 1;
if(i % pri[j])
sumu[i * pri[j]] = -sumu[i];
else {
sumu[i * pri[j]] = 0;
break;
}
}
}
for(int i = 2; i <= n; i++)
sumu[i] += sumu[i - 1];
}
ll Sum(int n, int m) {
ll res = 0;
int N = min(n, m);
for(int l = 1, r; l <= N; l = r + 1) {
r = min(n / (n / l), m / (m / l));
res += 1ll * (sumu[r] - sumu[l - 1]) * (n / l) * (m / l);
}
return res;
}
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
#endif // Yinku
sieve(MAXN);
int a, b, c, d;
while(~scanf("%d%d%d%d", &a, &b, &c, &d)) {
printf("%lld\n", Sum(b, d) - Sum(b, c - 1) - Sum(a - 1, d) + Sum(a - 1, c - 1));
}
} C - Contest Setting
这个逗比题怎么弄这么久。要求不可重，恰好k种，去重之后n²dp就可以了。
#include
using namespace std;
typedef long long ll;
const int MOD=998244353;
int a[1005];
int cnt[1005];
int dp[1005][1005];
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
#endif // Yinku
int n,k;
scanf("%d%d",&n,&k);
for(int i=1;i<=n;++i){
scanf("%d",&a[i]);
cnt[a[i]]++;
}
sort(a+1,a+1+n);
int t=unique(a+1,a+1+n)-(a+1);
dp[0][0]=1;
for(int i=1;i<=t;++i){
dp[i][0]=1;
for(int j=1;j<=min(i,k);++j)
dp[i][j]=(1ll*dp[i-1][j-1]*cnt[a[i]]%MOD+dp[i-1][j])%MOD;
}
printf("%d\n",dp[t][k]);
} L -Liars
应该是从大到小暴力枚举。
#include
using namespace std;
typedef long long ll;
int a[1005];
int b[1005];
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
#endif // Yinku
int n, ans = -1;
scanf("%d", &n);
for(int i = 1; i <= n; ++i) {
scanf("%d%d", &a[i], &b[i]);
}
for(int i = n; i >= 0; --i) {
int cnt = 0;
for(int j = 1; j <= n; ++j) {
if(a[j] <= i && i <= b[j])
++cnt;
}
if(cnt == i) {
ans = cnt;
break;
}
}
printf("%d\n", ans);
} J - Time Limits
#include
using namespace std;
typedef long long ll;
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
#endif // Yinku
int n, s;
scanf("%d%d", &n, &s);
int maxtime = 0;
for(int i = 1; i <= n; ++i) {
int t;
scanf("%d", &t);
maxtime = max(maxtime, t);
}
printf("%d\n", ((maxtime * s) + 999) / 1000);
}