链接：http://acm.bnu.edu.cn/v3/contest_show.php?cid=6865#info
2.5H写了三道题，后面撸C题KMP一直TLE T.T三题就三题吧……
A.Easy Math
灵感突现，想到“无理数与任何数的和都不可能为整数”，居然1A，真是勇气的试练。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <iomanip> 6 #include <cmath> 7 #include <map> 8 #include <vector> 9 #include <string> 10 #include <queue> 11 #include <set> 12 #include <algorithm> 13 14 using namespace std; 15 16 int n; 17 int tmp; 18 19 int main() { 20 while(~scanf("%d", &n)) { 21 int flag = true; 22 for(int i = 0; i < n; i++) { 23 scanf("%d", &tmp); 24 double sq = sqrt(tmp); 25 if(int(sq) != sq) { 26 flag = false; 27 } 28 } 29 if(flag) { 30 printf("Yes "); 31 } 32 else { 33 printf("No "); 34 } 35 } 36 }
B.Carries
计算两两和有多少个进位。卡在了mod取10^9的时候会爆int，无数次WA…
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <iomanip> 6 #include <cmath> 7 #include <map> 8 #include <vector> 9 #include <string> 10 #include <queue> 11 #include <set> 12 #include <algorithm> 13 14 using namespace std; 15 16 typedef long long LL; 17 const int maxn = 1000010; 18 int n; 19 int a[maxn]; 20 int h[maxn]; 21 22 int main() { 23 // freopen("in", "r", stdin); 24 while(~scanf("%d", &n)) { 25 memset(h, 0, sizeof(h)); 26 for(int i = 0; i < n; i++) { 27 scanf("%d", &a[i]); 28 } 29 LL ans = 0; 30 for(LL mod = 10; mod <= 1000000000; mod*=10) { //no more than 10^9 31 for(int i = 0; i < n; i++) { 32 h[i] = a[i] % mod; 33 } 34 sort(h, h+n); 35 int m = n - 1; 36 for(int i = 0; i < n; i++) { 37 while(i < m && h[i] + h[m] >= mod) { 38 m--; 39 } 40 ans += min(n-i-1, n-m-1); 41 } 42 } 43 printf("%lld ", ans); 44 } 45 }
E.Rectangle
和某微软笔试一样，确定一边暴力枚举另一边。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <iomanip> 6 #include <cmath> 7 #include <map> 8 #include <vector> 9 #include <string> 10 #include <queue> 11 #include <set> 12 #include <algorithm> 13 14 using namespace std; 15 16 typedef long long LL; 17 LL n, m, k; 18 19 int main() { 20 // freopen("in", "r", stdin); 21 while(~scanf("%lld %lld %lld", &n, &m, &k)) { 22 LL ans = 0; 23 k /= 2; 24 LL a, b; 25 for(LL i = 1; i <= n; i++) { 26 if(k < i) { 27 break; 28 } 29 LL tmp = min(k-i, m); 30 a = n - i + 1; 31 b = (m + (m - tmp + 1)) * tmp / 2; 32 ans += a * b; 33 } 34 printf("%lld ", ans); 35 } 36 }
补题ing……