結果
問題 |
No.896 友達以上恋人未満
|
ユーザー |
![]() |
提出日時 | 2018-03-02 18:21:46 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
MLE
(最新)
AC
(最初)
|
実行時間 | - |
コード長 | 1,183 bytes |
コンパイル時間 | 807 ms |
コンパイル使用メモリ | 67,460 KB |
最終ジャッジ日時 | 2025-01-05 08:57:34 |
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 5 MLE * 2 |
コンパイルメッセージ
main.cpp: In function ‘int main()’: main.cpp:25:44: warning: format ‘%d’ expects argument of type ‘int*’, but argument 2 has type ‘ll*’ {aka ‘long long int*’} [-Wformat=] 25 | for(int i = 0; i < m; i++) scanf("%d", x+i); | ~^ ~~~ | | | | int* ll* {aka long long int*} | %lld main.cpp:26:44: warning: format ‘%d’ expects argument of type ‘int*’, but argument 2 has type ‘ll*’ {aka ‘long long int*’} [-Wformat=] 26 | for(int i = 0; i < m; i++) scanf("%d", y+i); | ~^ ~~~ | | | | int* ll* {aka long long int*} | %lld main.cpp:35:44: warning: format ‘%d’ expects argument of type ‘int*’, but argument 2 has type ‘ll*’ {aka ‘long long int*’} [-Wformat=] 35 | for(int i = 0; i < m; i++) scanf("%d", a+i); | ~^ ~~~ | | | | int* ll* {aka long long int*} | %lld main.cpp:36:44: warning: format ‘%d’ expects argument of type ‘int*’, but argument 2 has type ‘ll*’ {aka ‘long long int*’} [-Wformat=] 36 | for(int i = 0; i < m; i++) scanf("%d", b+i); | ~^ ~~~ | | | | int* ll* {aka long long int*} | %lld main.cpp:24:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with at
ソースコード
// naive2 #include <cstdio> #include <cassert> #include <iostream> #include <cstring> using namespace std; typedef long long ll; #define N (1<<24) #define M 1000 int m, n, mx, ax, my, ay, mod; ll x[M], y[M], z[N], a[M], b[M], dp[N]; ll solve2(ll a){ if(a>mod) return 0; if(dp[a]>=0) return dp[a]; ll res = 0; for(ll i = a; i < mod; i+=a){ res += z[i]; } return dp[a] = res; } int main(){ scanf("%d%d%d%d%d%d%d",&m,&n,&mx,&ax,&my,&ay,&mod); for(int i = 0; i < m; i++) scanf("%d", x+i); for(int i = 0; i < m; i++) scanf("%d", y+i); for(int i = 0; i < m; i++) z[x[i]] += y[i]; ll xx = x[m-1], yy = y[m-1]; for(int i = m; i < n; i++){ xx = (xx*mx+(ll)ax)&(mod-1); yy = (yy*my+(ll)ay)&(mod-1); z[xx] += yy; } for(int i = 0; i < m; i++) scanf("%d", a+i); for(int i = 0; i < m; i++) scanf("%d", b+i); memset(dp, -1, sizeof(dp)); ll res = 0; for(int i = 0; i < m; i++){ ll r = solve2(a[i])-solve2(a[i]*b[i]); printf("%lld\n", r); res^=r; } ll aa = a[m-1], bb = b[m-1]; for(int i = m; i < n; i++){ aa = ((aa*mx+(ll)ax+mod-1)&(mod-1))+1; bb = ((bb*my+(ll)ay+mod-1)&(mod-1))+1; res^=solve2(aa)-solve2(aa*bb); } printf("%lld\n", res); return 0; }