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二分代价，由于序列是单增的，我们使前面一个数相对取最小，这样后面的数变化的值也能相对较小wa了好多次，发现上限少取了个0
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 #include<queue> 7 #include<map> 8 using namespace std; 9 #define MOD 1000000007 10 const int INF=0x3f3f3f3f; 11 const double eps=1e-5; 12 typedef long long ll; 13 #define cl(a) memset(a,0,sizeof(a)) 14 #define ts printf("***** "); 15 const int MAXN=100005; 16 int n,m,tt; 17 int a[MAXN],b[MAXN]; 18 bool check(int mid) 19 { 20 for(int i=1;i<=n;i++) 21 { 22 b[i]=a[i]; 23 } 24 for(int i=1;i<=n;i++) 25 { 26 if(i==1) 27 { 28 b[i]-=mid; 29 continue; 30 } 31 if(b[i]<=b[i-1]) 32 { 33 if((b[i-1]+1)-b[i]<=mid) 34 b[i]=b[i-1]+1; 35 else 36 { 37 return 0; 38 } 39 40 } 41 if(b[i]-mid>b[i-1]) 42 { 43 b[i]=b[i]-mid; 44 } 45 else 46 { 47 b[i]=b[i-1]+1; 48 } 49 } 50 return 1; 51 52 } 53 int main() 54 { 55 int i,j,k; 56 #ifndef ONLINE_JUDGE 57 freopen("1.in","r",stdin); 58 #endif 59 int ca=1; 60 scanf("%d",&tt); 61 while(tt--) 62 { 63 printf("Case #%d: ",ca++); 64 scanf("%d",&n); 65 for(i=1;i<=n;i++) scanf("%d",a+i); 66 int l=0; 67 int r=1000005; 68 int ans=0; 69 while(l<=r) 70 { 71 int mid=(l+r)>>1; 72 if(check(mid)) 73 { 74 ans=mid; 75 r=mid-1; 76 } 77 else 78 l=mid+1; 79 } 80 printf("%d ",ans); 81 } 82 }