http://poj.org/problem?id=2566；
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Bound Found
Description Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration (that must be going through a defiant phase: "But I want to use feet, not meters!"). Each signal seems to come in two parts: a sequence of n integer values and a non-negative integer t. We'll not go into details, but researchers found out that a signal encodes two integer values. These can be found as the lower and upper bound of a subrange of the sequence whose absolute value of its sum is closest to t.
You are given the sequence of n integers and the non-negative target t. You are to find a non-empty range of the sequence (i.e. a continuous subsequence) and output its lower index l and its upper index u. The absolute value of the sum of the values of the sequence from the l-th to the u-th element (inclusive) must be at least as close to t as the absolute value of the sum of any other non-empty range. Input The input file contains several test cases. Each test case starts with two numbers n and k. Input is terminated by n=k=0. Otherwise, 1<=n<=100000 and there follow n integers with absolute values <=10000 which constitute the sequence. Then follow k queries for this sequence. Each query is a target t with 0<=t<=1000000000.
Output For each query output 3 numbers on a line: some closest absolute sum and the lower and upper indices of some range where this absolute sum is achieved. Possible indices start with 1 and go up to n.
Sample Input 5 1 -10 -5 0 5 10 3 10 2 -9 8 -7 6 -5 4 -3 2 -1 0 5 11 15 2 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 15 100 0 0 Sample Output 5 4 4 5 2 8 9 1 1 15 1 15 15 1 15 Source |
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题意：找连续的串，求和绝对值与所给数字最接近的串。
思路：先求下标为1的到其他下的串的值，也就是前缀和；这样可以在O(1)的时间内求出各个串的和.比如S1(1,1),S3(1,3);
那么S3-S1就是S（2，3）；
然后对上面所求的前缀和按从小到大排序。（因为取的是绝对值所以abs(Sn-Sk)==abs(Sk-Sn)）;
这样就可以用尺取法去做了。复杂度为O(n);
还可以用二分取找值，复杂度为O(n*log(n));
1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<math.h> 6 #include<vector> 7 #include<map> 8 #include<stack> 9 int cmp(const void*p,const void*q); 10 typedef struct pp 11 { 12 int cost ; 13 int x; 14 int y; 15 } ss; 16 const int N=1<<30; 17 const int M=1e6; 18 int a[M]; 19 int b[M]; 20 ss dp[M]; 21 using namespace std; 22 int main(void) 23 { 24 int n,i,j,k,p,q; 25 while(scanf("%d %d",&p,&q),p!=0&&q!=0) 26 { 27 for(i=1; i<=p; i++) 28 { 29 scanf("%d",&a[i]); 30 } 31 dp[1].cost=a[1]; 32 dp[1].x=1; 33 dp[1].y=1; 34 for(i=2; i<=p; i++) 35 { 36 dp[i].cost=a[i]+dp[i-1].cost; 37 dp[i].x=1; 38 dp[i].y=i; 39 } 40 qsort(dp+1,p,sizeof(ss),cmp); 41 while(q--) 42 { 43 scanf("%d",&k); 44 int ll=1; 45 int rr=1; 46 int minn=N; 47 int lb; 48 int rb; 49 int co; 50 for(i=1; i<=p; i++) 51 { 52 if(abs(abs(dp[i].cost)-k)<minn) 53 { 54 minn=abs(abs(dp[i].cost)-k); 55 lb=1; 56 rb=dp[i].y; 57 co=abs(abs(dp[i].cost)); 58 } 59 } 60 while(rr<p) 61 { 62 while(rr<=p&&abs(dp[rr].cost-dp[ll].cost)<=k) 63 { 64 rr++; 65 } 66 if(rr==p+1) 67 { 68 rr--; 69 } 70 if(abs(dp[rr].cost-dp[ll].cost)>k&&rr-ll>1) 71 { 72 int nx=abs(abs(dp[rr].cost-dp[ll].cost)-k); 73 int ny=abs(abs(dp[rr-1].cost-dp[ll].cost)-k); 74 int kl=rr; 75 if(ny<nx) 76 { 77 kl--; 78 } 79 if(abs(abs(dp[kl].cost-dp[ll].cost)-k)<minn) 80 { 81 lb=min(dp[kl].y,dp[ll].y)+1; 82 rb=max(dp[kl].y,dp[ll].y); 83 minn=abs(abs(dp[kl].cost-dp[ll].cost)-k); 84 co=abs(dp[kl].cost-dp[ll].cost); 85 } 86 } 87 else if(abs(dp[rr].cost-dp[ll].cost)>k&&rr-ll==1) 88 { 89 if(abs(abs(dp[rr].cost-dp[ll].cost)-k)<minn) 90 { 91 lb=min(dp[rr].y,dp[ll].y)+1; 92 rb=max(dp[rr].y,dp[ll].y); 93 minn=abs(abs(dp[rr].cost-dp[ll].cost)-k); 94 co=abs(dp[rr].cost-dp[ll].cost); 95 } 96 97 } 98 else if(abs(dp[rr].cost-dp[ll].cost)<=k) 99 { 100 if(abs(abs(dp[rr].cost-dp[ll].cost)-k)<minn) 101 { 102 lb=min(dp[rr].y,dp[ll].y)+1; 103 rb=max(dp[rr].y,dp[ll].y); 104 minn=abs(abs(dp[rr].cost-dp[ll].cost)-k); 105 co=abs(dp[rr].cost-dp[ll].cost); 106 } 107 108 } 109 ll++; 110 } 111 printf("%d %d %d ",co,lb,rb); 112 } 113 } 114 } 115 int cmp(const void*p,const void*q) 116 { 117 ss*ww=(ss*)p; 118 ss*w=(ss*)q; 119 if(ww->cost==w->cost) 120 { 121 return ww->y-w->y; 122 } 123 return ww->cost-w->cost; 124 }
二分
1 #include<stdio.h> 2 #include<algorithm> 3 #include<stdlib.h> 4 #include<string.h> 5 #include<algorithm> 6 #include<iostream> 7 #include<map> 8 const int N=1000000; 9 int cmp(const void*p,const void*q); 10 int er(int n,int m,int t,int c); 11 typedef struct pp 12 { 13 int cost; 14 int le; 15 int ri; 16 } ss;//结构体记录前缀数组和和其左右端点 17 ss a[N]; 18 int b[N],c[N]; 19 int vv; 20 using namespace std; 21 int main(void) 22 { 23 int n,i,j,k,p,q; 24 while(scanf("%d %d",&p,&q),p!=0&&q!=0) 25 { 26 for(i=1; i<=p; i++) 27 scanf("%d",&b[i]); 28 int cnt=1; 29 a[cnt].cost=b[1]; 30 a[cnt].le=1; 31 a[cnt].ri=1; 32 cnt++; 33 for(i=2; i<=p; i++) 34 { 35 a[cnt].cost=a[cnt-1].cost+b[i]; 36 a[cnt].le=1; 37 a[cnt].ri=i; 38 cnt++; 39 } 40 qsort(a+1,cnt-1,sizeof(ss),cmp); 41 /*for(i=1;i<=p;i++) 42 { 43 printf("%d ",a[i].cost); 44 printf("000 "); 45 printf("%d ",a[i].ri); 46 }*/ 47 for(i=1; i<=q; i++) 48 { 49 cin>>b[i]; 50 } 51 for(int pq=1; pq<=q; pq++) 52 { 53 k=b[pq]; 54 int minn=1<<30; 55 int ll=0; 56 int rr=0; 57 int uk=0; 58 for(j=1; j<=p; j++) 59 { 60 if(abs(abs(a[j].cost)-k)<minn) 61 { 62 minn=abs(abs(a[j].cost)-k); 63 uk=abs(a[j].cost); 64 ll=1; 65 rr=a[j].ri; 66 } 67 }//首先把所有前缀过一遍更新最小，因为二分时找不到 68 for(j=1; j<p; j++) 69 { 70 int kk=er(j+1,p,k,j); 71 vv=j; 72 if(kk==p) 73 {if(kk-vv>1)//这里需要判断下，kk-vv>1才能用下面的比较。 74 {int uu=abs(abs(a[kk-1].cost-a[j].cost)-k); 75 int w=abs(abs(a[kk].cost-a[j].cost)-k); 76 if(uu<w) 77 { 78 kk--; 79 } 80 int ww=abs(abs(a[kk].cost-a[j].cost)-k); 81 if(ww<minn) 82 { 83 minn=ww; 84 ll=min(a[kk].ri,a[j].ri)+1; 85 rr=max(a[kk].ri,a[j].ri); 86 uk=abs(a[kk].cost-a[j].cost); 87 }} 88 else if(kk-vv==1)//kk-vv正好为1时表明kk就为所找，不能再比较kk-1，因为kk-1为空值 89 { int ww=abs(abs(a[kk].cost-a[j].cost)-k); 90 if(ww<minn) 91 { 92 minn=ww; 93 ll=min(a[kk].ri,a[j].ri)+1; 94 rr=max(a[kk].ri,a[j].ri); 95 uk=abs(a[kk].cost-a[j].cost); 96 } 97 } 98 } 99 else 100 { 101 102 int uu=abs(abs(a[kk-1].cost-a[j].cost)-k); 103 int w=abs(abs(a[kk].cost-a[j].cost)-k); 104 if(uu<w) 105 { 106 kk--; 107 } 108 if(kk==j) 109 { 110 kk++; 111 } 112 int ww=abs(abs(a[kk].cost-a[j].cost)-k); 113 if(ww<minn) 114 { 115 minn=ww; 116 ll=min(a[kk].ri,a[j].ri)+1; 117 rr=max(a[kk].ri,a[j].ri); 118 uk=abs(a[kk].cost-a[j].cost); 119 120 } 121 } 122 123 } 124 printf("%d %d %d ",uk,ll,rr); 125 } 126 } 127 } 128 int cmp(const void*p,const void*q)//二分查找 129 { 130 ss*w=(ss*)p; 131 ss*ww=(ss*)q; 132 return w->cost-ww->cost; 133 } 134 int er(int n,int m,int k,int c) 135 { 136 int y=(n+m)/2; 137 if(abs(a[y].cost-a[c].cost)==k) 138 { 139 return y; 140 } 141 else if(n==m) 142 { 143 return m; 144 } 145 else if(abs(a[y].cost-a[c].cost)>k&&abs(a[y-1].cost-a[c].cost)<k) 146 { 147 return y; 148 } 149 else if(abs(a[y].cost-a[c].cost)<k&&abs(a[y-1].cost-a[c].cost)<k) 150 { 151 return er(y+1,m,k,c); 152 } 153 else return er(n,y-1,k,c); 154 }