AcWing题库思路一眼二分典中典先排个序再用lower_bound和upper_bound维护相同的数的左界和右界就好了注意特判无解Code#include

bits/stdc.h

pos2upper_bound(a,an,x)-a-1;if((!check(pos1)||!check(pos2))||(pos1pos2))

cout-1

main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int

__1;//cin__;init();while(__--)solve();return

}790.

AcWing题库思路注意到答案具有单调性因此二分即可Code#include

bits/stdc.h

l-10000.0,r10000.0;while(abs(r-l)eps){double

mid(lr)/2;if(check(mid)){ansmid;rmid;}else

lmid;}coutfixedsetprecision(6)ans\n;

void

main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int

__1;//cin__;init();while(__--)solve();return

}AcWing

AcWing思路大一学弟也会写的前缀和板子Code#include

bits/stdc.h

cina[i],sum[i]sum[i-1]a[i];while(m--){cinlr;coutsum[r]-sum[l-1]\n;}

void

main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int

__1;//cin__;init();while(__--)solve();return

}796.

AcWing题库同样是大一学弟也会的二维前缀和但是我可能不太会嘻Code#include

bits/stdc.h

sum[i][j]sum[i-1][j]sum[i][j-1]a[i][j]-sum[i-1][j-1];}while(q--){cinx1y1x2y2;coutsum[x2][y2]-sum[x1-1][y2]-sum[x2][y1-1]sum[x1-1][y1-1]\n;}

void

main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int

__1;//cin__;init();while(__--)solve();return

}730.

AcWing题库思路答案具有二分性可以直接二分然后在check函数里去模拟这个过程如果中间存在能量值0的情况就false否则如果出现大于maxH的情况那么剩下的只会得到因此直接true就好了Code#include

bits/stdc.h

i0;in;i){if(h[i1]res){res-(h[i1]-res);}else{res(res-h[i1]);}if(resmx)

return

midlr1;if(check(mid)){ansmid;rmid-1;}else

void

main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int

__1;//cin__;init();while(__--)solve();return

}AcWing

AcWing思路四指针枚举想到先把两个指针的结果哈希一下然后再去枚举两个指针一个指针的复杂度是sqrt(n)两个就是O(n)的了Code#include

bits/stdc.h

j0;i*ij*jn;j){v[i*ij*j]{i,j};}}for(int

i0;i*in;i){for(int

j0;i*ij*jn;j){if(v.count(n-i*i-j*j)){couti

v[n-i*i-j*j].second

v[n-i*i-j*j].first\n;return;}}}

void

main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int

__1;//cin__;init();while(__--)solve();return

}1227.

AcWing题库思路直接去二分边长然后去check函数计算能有多少巧克力块就行Code#include

bits/stdc.h

i1;in;i){res(p[i].h/x)*(p[i].w/x);}return

resk;

midlr1;if(check(mid)){ansmid;lmid1;}else

rmid-1;}coutans\n;

main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int

__1;//cin__;init();while(__--)solve();return

}99.

AcWing题库思路数据范围都比较小因此可以直接求个二维前缀和然后求二维差分维护最大值即可#include

bits/stdc.h

solve(){cinnr;rmin(r,5001);for(int

i1;in;i){cinxyw;x,y;a[x][y]w;}for(int

i1;i5001;i){for(int

a[i][j]a[i-1][j]a[i][j-1]-a[i-1][j-1];}int

ans-1e9;for(int

Sa[x4][y4]-a[x2-1][y2]-a[x3][y3-1]a[x1-1][y1-1];ansmax(ans,S);}}coutans\n;

void

main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int

__1;//cin__;init();while(__--)solve();return

}1230.

AcWing题库思路很典直接求前缀和然后对前缀和取模k两个模为0的点就可以作为k倍区间端点然后用动态map维护即可也可以算C(n,2)Code#include

bits/stdc.h

cina[i],sum[i]sum[i-1]a[i],sum[i]%k;int

ans0;for(int

i1;in;i){ansmp[sum[i]];mp[sum[i]];}coutansmp[0]\n;

void

main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int

__1;//cin__;init();while(__--)solve();return