AtCoder Beginner Contest 103
AtCoder Beginner Contest 103
attack
发布于 2018-07-27 15:27:20
发布于 2018-07-27 15:27:20
刚刚天真的跑去打codechef，才发现那是IOI模拟赛qwq。
atc是比赛还剩40min结束的时候才打的，就做了前三个题
T1
zz模拟
#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int a[5];
int main() {
a[1] = read(); a[2] = read(); a[3] = read();
sort(a + 1, a + 4);
printf("%d", abs(a[2] - a[1]) + abs(a[3] - a[2]));
return 0;
}T2
zz模拟
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
string a, b;
int main() {
cin >> a >> b;
int N = a.length();
if(N != b.length()) {puts("No"); return 0;}
//for(int i = 0; i < N; i++) a[i + N] = a[i];
a = a + a;
//cout << a << endl << b;
if(a.find(b) != string::npos) puts("Yes");
else puts("No");
return 0;
}T3
这个就比较厉害了。
首先我们发现，对于每个$a_i$，我们都可以构造一个数使得$x \pmod {a_i} = a_i - 1$
那么输出$\sum_{i = 1}^n a_i - 1$就行了
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
//#define int long long
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N;
int a[MAXN], sum = 0;
int check(int val) {
int ret = 0;
for(int i = 1; i <= N; i++)
ret += val % a[i];
return ret;
}
main() {
N = read();
for(int i = 1; i <= N; i++)
a[i] = read(), sum += a[i] - 1;
printf("%d", sum);
return 0;
}T4
首先按照套路按照右端点排序
然后按照从左到右的顺序依次在右端点切就行了
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#define MP(x, y) make_pair(x, y)
#define Pair pair<int, int>
//#define int long long
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, M;
Pair P[MAXN];
main() {
N = read(); M = read();
for(int i = 1; i <= M; i++) {
int x = read(), y = read();
P[i] = MP(y, x);
}
int ans = 0, last = -1;
sort(P + 1, P + M + 1);
for(int i = 1; i <= M;) {
int j = i + 1;
while(j <= M && P[i].first == P[j].first) j++;
if(P[j - 1].second >= last)
ans++, = P[j - 1].first;
i = j;
}
printf("%d\n", ans);
return 0;
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原始发表：2018-07-21 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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