传送门
题意： 有n层具有颜色的涂层，其每层的颜色由Ri, Ci,Bi三个指标对应6个长度的16进制数表示的。现问你 [li, ri]区间涂层的颜色重叠在一起是什么颜色。重叠规则如下：
* 若第i层的 wi==1，那么它只会显示自己的颜色(Ri,Gi,Bi)。
* 若第i层的 wi==2，那么它的颜色就是自身与先前层次的颜色的混和(min(Rp+Ri,255), min(Gp+Gi,255), min(Bp+Bi,255)，当然颜色会有个上限255。
针对 p 次询问回答每个区间最后的颜色是什么。
思路：不难看出这就是在求 [li, ri]的颜色区间和，只不过换成了16进制数；且当出现 '1'时就会阻断颜色的传递，即所求颜色为 [max(pos, l), r]区间内的和，pos为最靠近 r 的前方 '1' 位置。
赛中还写了16->10和10->16进制之间转换的自定义函数，但是直接T到想吐！！！最后甚至只用到10->16的转换并于map初始化标记，还利用了快读和getchar()，依旧T到怀疑人生.......
代码实现：
#include<bitsdc++.h>
#define LL long long
#define me(ar) memset(ar, 0, sizeof ar)
using namespace std;
const int N = 1e5 + 5;inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;
}int T = 1;int n, p, l, r, sa, sb, sc;
string s[N];
/ar s[N][10];
unordered_map<string, int> mp;
unordered_map<int, string> mp2;
int m[N], pos[N], a[N], b[N], c[N];
int pa[N], pb[N], pc[N], arr[N], num[N];
char h[16]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};void work1(int x){int cnt = 0, w = x;string hh;while(x>0){num[cnt++]=x%16;x /= 16;}if(cnt==1) hh += "0";for(int i = cnt-1; ~i; i --){int m = num[i];hh += h[m];}mp[hh] = w;mp2[w] = hh;
}void Inint(){for(int i = 1; i <= 255; i ++){work1(i);}
}signed main()
{Inint();T = read();while(T --){n = read(); p = read();for(int i = 1; i <= n; i ++){m[i] = read();
// scanf("%s", s[i]);
// getchar();for(int j = 1; j <= 6; j ++){s[i] += getchar();}
// getchar();
// cout << "s: " << s[i] << endl;string ss1, ss2, ss3;ss1 += s[i][0]; ss1 += s[i][1];a[i] = mp[ss1];ss2 += s[i][2]; ss2 += s[i][3];b[i] = mp[ss2];ss3 += s[i][4]; ss3 += s[i][5];c[i] = mp[ss3];
// cout << "a: " << a[i] << " b: " << b[i] << " c: " << c[i] << endl;pa[i] = pa[i-1]+a[i];pb[i] = pb[i-1]+b[i];pc[i] = pc[i-1]+c[i];if(m[i]==1) pos[i] = i;else pos[i] = pos[i-1];}while(p --){l = read(); r = read();if(pos[r]==r){int le = s[r].size();for(int i = 0; i < le; i ++)printf("%c", s[r][i]);}else if(pos[r]>l){int ll = pos[r];sa = pa[r]-pa[ll-1], sb = pb[r]-pb[ll-1], sc = pc[r]-pc[ll-1];sa = min(sa, 255), sb = min(sb, 255), sc = min(sc, 255);int la = mp2[sa].size(), lb = mp2[sb].size(), lc = mp2[sc].size();for(int i = 0; i < la; i ++)printf("%c", mp2[sa][i]);for(int i = 0; i < lb; i ++)printf("%c", mp2[sb][i]);for(int i = 0; i < lc; i ++)printf("%c", mp2[sc][i]);}else {sa = pa[r]-pa[l-1], sb = pb[r]-pb[l-1], sc = pc[r]-pc[l-1];sa = min(sa, 255), sb = min(sb, 255), sc = min(sc, 255);int la = mp2[sa].size(), lb = mp2[sb].size(), lc = mp2[sc].size();for(int i = 0; i < la; i ++)printf("%c", mp2[sa][i]);for(int i = 0; i < lb; i ++)printf("%c", mp2[sb][i]);for(int i = 0; i < lc; i ++)printf("%c", mp2[sc][i]);}puts("");}}return 0;
}
呜呜呜~赛后深刻反思，不该一直纠结一个思路和一份代码硬磕，及时转换思想很重要！！
其实此题在进制之间的转换只涉及到两位数，每次利用的时候临时计算一下反而比规范的自定义函数更省时；再者去掉自定义函数后甚至可以利用纯C语言编写最简洁的程序，没有了string和map肯定更省时。
AC代码：
#include<iostream>
#include<algorithm>
#include<unordered_map>
#define LL long long
#define me(ar) memset(ar, 0, sizeof ar)
using namespace std;
const int N = 1e5 + 5;inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;
}int T = 1;int n, p, l, r, a, b, c, sa, sb, sc;
char s[10];
int m[N], pos[N], pa[N], pb[N], pc[N];
char h[18]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};int matche(char o){if('0'<=o&&o<='9') return o-'0';else return o-'A'+10;
}signed main()
{T = read();while(T --){n = read(); p = read();for(int i = 1; i <= n; i ++){m[i] = read(); scanf("%s", s);a = matche(s[0])*16+matche(s[1]);b = matche(s[2])*16+matche(s[3]);c = matche(s[4])*16+matche(s[5]);pa[i] = pa[i-1]+a; pb[i] = pb[i-1]+b; pc[i] = pc[i-1]+c;if(m[i]==1) pos[i] = i;else pos[i] = pos[i-1];}while(p --){l = read(); r = read();if(pos[r]>l) l = pos[r];sa = pa[r]-pa[l-1], sb = pb[r]-pb[l-1], sc = pc[r]-pc[l-1];sa = min(sa, 255), sb = min(sb, 255), sc = min(sc, 255);printf("%c%c", h[sa/16], h[sa%16]);printf("%c%c", h[sb/16], h[sb%16]);printf("%c%c", h[sc/16], h[sc%16]);puts("");}}return 0;
}