题意：给你一个二叉树，根节点为1，子节点为父节点的2倍和2倍+1，从根节点开始依次向下走k层，问如何走使得将路径上的数进行加减最终结果得到n。
联想到二进制。
思路和这个差不多吧：http://blog.csdn.net/u013068502/article/details/50094561
#include <set>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
#define mem(x,y) memset(x, y, sizeof(x))
#define lson l,m,rt << 1
#define rson m+1,r,rt << 1 | 1
? a : gcd(b, a % b);}
int lcm(int a, int b){return a / gcd(a, b) * b;}
int main()
{
int T;
scanf("%d", &T);
; cas <= T; cas++)
{
;
scanf("%d%d", &n, &k);
== )
{
n--;
flag = ;
}
LL all = ( << (k + )) - ;
LL res = (all - n) >> ;
printf("Case #%d:\n", cas);
; i < k; i++)
{
LL temp = << i;
)) printf( : temp);
) printf("%I64d -\n", temp);
else printf("%I64d +\n", temp);
res >>= ;
}
}
;
}
题意：给你一个字符串,找最近的相同字符
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
#define mem(x,y) memset(x, y, sizeof(x))
#define lson l,m,rt << 1
#define rson m+1,r,rt << 1 | 1
? a : gcd(b, a % b);}
int lcm(int a, int b){return a / gcd(a, b) * b;}
const int INF = 0x3f3f3f3f;
];
int main()
{
int T;
scanf("%d", &T);
getchar();
; cas <= T; cas++)
{
scanf("%s", s);
int len = strlen(s);
int ans = INF;
; i < len; i++)
{
; j < len; j++)
{
if(s[i] == s[j])
{
ans = min(ans, abs(i - j));
}
}
}
printf( : ans);
}
;
}
5583 Kingdom of Black and White
题意：给你一个01字符串,要求最多改变一个字符(即0->1或1->0),使相邻相同字符的平方和最大
漏了考虑，改变中间一个区间之后，两个到三个区间变成相同字符的情况。WA了。
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
#define mem(x,y) memset(x, y, sizeof(x))
#define lson l,m,rt << 1
#define rson m+1,r,rt << 1 | 1
? a : gcd(b, a % b);}
int lcm(int a, int b){return a / gcd(a, b) * b;}
;
int s[maxn];
LL len[maxn];
int main()
{
int T;
scanf("%d", &T);
getchar();
; cas <= T; cas++)
{
mem(s, -);
char ch;
;
while(~scanf("%c", &ch))
{
if(ch == '\n') break;
s[slen++] = ch - ';
}
, cur = , cnt = ;
while(cur <= slen)
{
])
{
len[cnt++] = cur - temp;
temp = cur;
}
cur++;
}
LL ans = ;
; i < cnt; i++)
{
ans += len[i] * len[i];
}
LL rec = ans;
; i < cnt; i++)
{
) break;
] == && i != )
{
ans = max(ans,
rec - len[i - ] * len[i - ] - len[i - ] * len[i - ] - len[i] * len[i]
+ (len[i - ] + len[i - ] + len[i]) * (len[i - ] + len[i - ] + len[i]));
}
])
{
ans = max(ans, rec + * (len[i - ] - len[i] + ));
}
])
{
ans = max(ans, rec + * (len[i] - len[i - ] + ));
}
}
printf("Case #%d: %I64d\n", cas, ans);
}
;
}
题意：有一只青蛙,它从起点(x,y)出发,每次它会走 LCM(x,y) 步到达点 (x+LCM(x,y),y) 或点 (x,y+LCM(x,y)) ,最终,它会到达点(ex,ey),现给你终点(ex,ey),要你求出它的起点有多少种可能。
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
#define mem(x,y) memset(x, y, sizeof(x))
#define lson l,m,rt << 1
#define rson m+1,r,rt << 1 | 1
? a : gcd(b, a % b);}
int lcm(int a, int b){return a / gcd(a, b) * b;}
;
int main()
{
int T, fx, fy;
scanf("%d", &T);
; cas <= T; cas++)
{
scanf("%d%d", &fx, &fy);
LL ans = ;
)
{
int g = gcd(fx, fy);
if(fy > fx) swap(fx, fy);
) break;
int m1 = fx / (fy + g);
int m2 = fy / g;
fx = m1 * g;
fy = m2 * g;
ans++;
}
printf();
}
;
}