CF2259problemE也是这个思想https://www.luogu.com.cn/problem/T700045题面https://codeforces.com/contest/2259/problem/E由约束推出可行域再检查可行域是否满足所有条件根据所给数组 推 构造数组的某些区间的限制条件也就是一些Forbidden区间类似题目https://codeforces.com/contest/2262/problem/A1Codevoidsolve(){intn;cinn;vectorinta(n1,0);for(inti1;in;i)cina[i];vectorintdif(n2,0);//Experience差分数组多开一点//如果要处理的是 [l,r]//右端点r缩到 n-1 , dif[r1]-- 就会处理错成 dif[n]--//Wrong wrong wrong//Right rmin(n,...) dif[n1]--才会处理对for(inti1;in;i){intxa[i];if(x0)continue;intlmax(1ll,i-x1);intrmin(n,ix-1);// if(lr)// {dif[l];dif[r1]--;// }}for(inti1;in;i)dif[i]dif[i-1];vectorintans(n1,1);for(inti1;in;i){if(dif[i]0)ans[i]0;}for(inti1;in;i){intxa[i];if(x-1)continue;boolf0;intli-x;if(l1)if(ans[l])f1;intrix;if(rn)if(ans[r])f1;if(!f){cout-1\n;return;}}for(inti1;in;i)coutans[i];cout\n;return;}Experience 差分数组开多一点右端点rmin(n,…);这样缩才能处理对