UVA 13154
https://gist.github.com/ss1h2a3tw/d87e3ebf67a08d631c6fd8fcabf2e08f
Codeforces 100800G
https://gist.github.com/ss1h2a3tw/77230a157243e4f557427d9e90c5fbcf
Shane Chen ss1h2a3tw
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| #include <cstdio> | |
| int main (){ | |
| char a[8*1024*1024]; | |
| a[sizeof(a)-1]=0; | |
| asm volatile (""::"g"(a):"memory"); | |
| printf("Success! %lld\n",(long long)sizeof(a)); | |
| } |
ss1h2a3tw
/ editorial.md
Created
January 13, 2017 21:10
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| /* | |
| Codeforces 100800G | |
| United Kingdom and Ireland Subregional Contest 2015 pG | |
| 今天有學生想把錢花完並且喝到一定單位的酒精 | |
| 給你酒的種類和價格和單位酒精含量 | |
| 然後輸出可不可行 可行的話輸出方法 |
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| /* | |
| UVA13154 Extreme XOR Sum | |
| 給你一串數列長度10^4 元素<10^9 | |
| 定義區間內XOR sum 為將左右鄰居兩兩相XOR | |
| 然後將結果做同樣操作 直到只剩一個元素即為XOR sum | |
| 會有30000個區間查詢 |
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| /* | |
| 給你一個Binary String長度 < 10^5 | |
| 定義成長操作為在兩兩間隔間插入 其XOR值 | |
| 查詢在經過 g < 10^4 成長後 區間內0,1的數量mod prime | |
| 思路 | |
| 首先查詢區間[a,b]可以以[0,b]-[0,a-1]來計算所以問題可以精簡成為單向成長 | |
| 那麼在一次成長過程中0增加的數量為相同的鄰居 1則反之 |