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常用的解題技巧：尺取法

尺取法：顧名思義，像尺子一樣取一段，借用挑戰書上面的話說，尺取法通常是對數組保存一對下標，即所選取的區間的左右端點，然后根據實際情況不斷地推進區間左右端點以得出答案。之所以需要掌握這個技巧，是因為尺取法比直接暴力枚舉區間效率高很多，尤其是數據量大的時候，所以尺取法是一種高效的枚舉區間的方法，一般用于求取有一定限制的區間個數或最短的區間等等。當然任何技巧都存在其不足的地方，有些情況下尺取法不可行，無法得出正確答案。

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使用尺取法時應清楚以下四點：

1、??什么情況下能使用尺取法? ?2、何時推進區間的端點？ 3、如何推進區間的端點？ 3、何時結束區間的枚舉？

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尺取法通常適用于選取區間有一定規律，或者說所選取的區間有一定的變化趨勢的情況，通俗地說，在對所選取區間進行判斷之后，我們可以明確如何進一步有方向地推進區間端點以求解滿足條件的區間，如果已經判斷了目前所選取的區間，但卻無法確定所要求解的區間如何進一步

得到根據其端點得到，那么尺取法便是不可行的。首先，明確題目所需要求解的量之后，區間左右端點一般從最整個數組的起點開始，之后判斷區間是否符合條件在根據實際情況變化區間的端點求解答案。

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以下是幾個經典的使用尺取法的例題，都是從挑戰書上引用的。（尺取法通常會需要對某些量進行預處理，以便能在使用時快速地判斷。）

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1、? Poj3061

題意：給定一個序列，找出最短的子序列長度，使得其和大于或等于S。

分析：首先，序列都是正數，如果一個區間其和大于等于S了，那么不需要在向后推進右端點了，因為其和也肯定大于等于S但長度更長，所以，當區間和小于S時右端點向右移動，和大于等于S時，左端點向右移動以進一步找到最短的區間，如果右端點移動到區間末尾其和還不大于等于S，

結束區間的枚舉。

這個題目區間和明顯是有趨勢的：單調變化，所以根據題目要求很容易求解，但是在使用之間需要對區間前綴和進行預處理計算。

代碼：

#include <cstdio> #include <algorithm> #include <cstring> #define MAX 100005 #define LL long long #define INF 0x3f3f3f3fusing namespace std; LL a[100010]; int n, t, ans = INF; LL sum, s;int main() {scanf("%d", &t);while (t--){scanf("%d %I64d", &n, &s);for (int i = 0; i < n; i++) scanf("%I64d", a+i);int st = 0, en = 0;ans = INF; sum = 0;while (1){while (en<n && sum<s) sum += a[en++];if (sum < s) break;ans = min(ans, en-st);sum -= a[st++];}if (ans == INF) ans = 0;printf("%d\n", ans);}return 0; }</code>

2、? poj3320

題意：一本書有P頁，每一頁都一個知識點，求去最少的連續頁數覆蓋所有的知識點。

分析：和上面的題一樣的思路，如果一個區間的子區間滿足條件，那么在區間推進到該處時，右端點會固定，左端點會向右移動到其子區間，且其子區間會是更短的，只是需要存儲所選取的區間的知識點的數量，那么使用map進行映射以快速判斷是否所選取的頁數是否覆蓋了所有的

知識點。

代碼：

#include <cstdio> #include <algorithm> #include <cstring> #include <set> #include <map> #define MAX 1000010 #define LL long long #define INF 0x3f3f3f3fusing namespace std; int a[MAX]; map <int, int> cnt; set <int> t; int p, ans = INF, st, en, sum;int main() {scanf("%d", &p);for (int i = 0; i < p; i++) scanf("%d", a+i), t.insert(a[i]);int num = t.size();while (1){while (en<p && sum<num)if (cnt[a[en++]]++ == 0) sum++;if (sum < num) break;ans = min(ans, en-st);if (--cnt[a[st++]] == 0) sum--;}printf("%d\n", ans);return 0; }

3、? poj2566

題意：給定一個數組和一個值t，求一個子區間使得其和的絕對值與t的差值最小，如果存在多個，任意解都可行。

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分析：明顯，借用第一題的思路，既然要找到一個子區間使得和最接近t的話，那么不斷地找比當前區間的和更大的區間，如果區間和已經大于等于t了，那么不需要在去找更大的區間了，因為其和與t的差值更大，然后區間左端點向右移動推進即可。所以，首先根據計算出所有的區間和，

排序之后按照上面的思路求解即可。

代碼：

#include <cstdio> #include <algorithm> #include <cstring> #define INF 0x3f3f3f3f #define LL long long #define MAX 100010 using namespace std;typedef pair<LL, int> p; LL a[MAX], t, ans, tmp, b; int n, k, l, u, st, en; p sum[MAX];LL myabs(LL x) {return x>=0? x:-x; }int main() {while (scanf("%d %d", &n, &k), n+k){sum[0] = p(0, 0);for (int i = 1; i <= n; i++){scanf("%I64d", a+i);sum[i] = p(sum[i-1].first+a[i], i);}sort(sum, sum+1+n);while (k--){scanf("%I64d", &t);tmp = INF; st = 0, en = 1;while(en <= n){b = sum[en].first-sum[st].first;if(myabs(t-b) < tmp){tmp = myabs(t-b);ans = b;l = sum[st].second; u = sum[en].second;}if(b > t) st++;else if(b < t) en++;else break;if(st == en) en++;}if (u < l) swap(u, l);printf("%I64d %d %d\n", ans, l+1, u);}}return 0; }

4、? poj2739&poj2100

題意：找到某一個區間使得區間內的數的和/平方和等于某一給定值k。

分析：很明顯了，幾乎之與上面的poj2566又是一樣的，當區間右端點不能再向右推進且區間和仍小于k的話就可以結束區間的枚舉了。

代碼：

poj2739；

#include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <utility> #include <queue> #define INF 0x3f3f3f3f #define LL long long using namespace std;int prime[] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053,2063,2069,2081,2083,2087,2089,2099,2111,2113,2129,2131,2137,2141,2143,2153,2161,2179,2203,2207,2213,2221,2237,2239,2243,2251,2267,2269,2273,2281,2287,2293,2297,2309,2311,2333,2339,2341,2347,2351,2357,2371,2377,2381,2383,2389,2393,2399,2411,2417,2423,2437,2441,2447,2459,2467,2473,2477,2503,2521,2531,2539,2543,2549,2551,2557,2579,2591,2593,2609,2617,2621,2633,2647,2657,2659,2663,2671,2677,2683,2687,2689,2693,2699,2707,2711,2713,2719,2729,2731,2741,2749,2753,2767,2777,2789,2791,2797,2801,2803,2819,2833,2837,2843,2851,2857,2861,2879,2887,2897,2903,2909,2917,2927,2939,2953,2957,2963,2969,2971,2999,3001,3011,3019,3023,3037,3041,3049,3061,3067,3079,3083,3089,3109,3119,3121,3137,3163,3167,3169,3181,3187,3191,3203,3209,3217,3221,3229,3251,3253,3257,3259,3271,3299,3301,3307,3313,3319,3323,3329,3331,3343,3347,3359,3361,3371,3373,3389,3391,3407,3413,3433,3449,3457,3461,3463,3467,3469,3491,3499,3511,3517,3527,3529,3533,3539,3541,3547,3557,3559,3571,3581,3583,3593,3607,3613,3617,3623,3631,3637,3643,3659,3671,3673,3677,3691,3697,3701,3709,3719,3727,3733,3739,3761,3767,3769,3779,3793,3797,3803,3821,3823,3833,3847,3851,3853,3863,3877,3881,3889,3907,3911,3917,3919,3923,3929,3931,3943,3947,3967,3989,4001,4003,4007,4013,4019,4021,4027,4049,4051,4057,4073,4079,4091,4093,4099,4111,4127,4129,4133,4139,4153,4157,4159,4177,4201,4211,4217,4219,4229,4231,4241,4243,4253,4259,4261,4271,4273,4283,4289,4297,4327,4337,4339,4349,4357,4363,4373,4391,4397,4409,4421,4423,4441,4447,4451,4457,4463,4481,4483,4493,4507,4513,4517,4519,4523,4547,4549,4561,4567,4583,4591,4597,4603,4621,4637,4639,4643,4649,4651,4657,4663,4673,4679,4691,4703,4721,4723,4729,4733,4751,4759,4783,4787,4789,4793,4799,4801,4813,4817,4831,4861,4871,4877,4889,4903,4909,4919,4931,4933,4937,4943,4951,4957,4967,4969,4973,4987,4993,4999,5003,5009,5011,5021,5023,5039,5051,5059,5077,5081,5087,5099,5101,5107,5113,5119,5147,5153,5167,5171,5179,5189,5197,5209,5227,5231,5233,5237,5261,5273,5279,5281,5297,5303,5309,5323,5333,5347,5351,5381,5387,5393,5399,5407,5413,5417,5419,5431,5437,5441,5443,5449,5471,5477,5479,5483,5501,5503,5507,5519,5521,5527,5531,5557,5563,5569,5573,5581,5591,5623,5639,5641,5647,5651,5653,5657,5659,5669,5683,5689,5693,5701,5711,5717,5737,5741,5743,5749,5779,5783,5791,5801,5807,5813,5821,5827,5839,5843,5849,5851,5857,5861,5867,5869,5879,5881,5897,5903,5923,5927,5939,5953,5981,5987,6007,6011,6029,6037,6043,6047,6053,6067,6073,6079,6089,6091,6101,6113,6121,6131,6133,6143,6151,6163,6173,6197,6199,6203,6211,6217,6221,6229,6247,6257,6263,6269,6271,6277,6287,6299,6301,6311,6317,6323,6329,6337,6343,6353,6359,6361,6367,6373,6379,6389,6397,6421,6427,6449,6451,6469,6473,6481,6491,6521,6529,6547,6551,6553,6563,6569,6571,6577,6581,6599,6607,6619,6637,6653,6659,6661,6673,6679,6689,6691,6701,6703,6709,6719,6733,6737,6761,6763,6779,6781,6791,6793,6803,6823,6827,6829,6833,6841,6857,6863,6869,6871,6883,6899,6907,6911,6917,6947,6949,6959,6961,6967,6971,6977,6983,6991,6997,7001,7013,7019,7027,7039,7043,7057,7069,7079,7103,7109,7121,7127,7129,7151,7159,7177,7187,7193,7207,7211,7213,7219,7229,7237,7243,7247,7253,7283,7297,7307,7309,7321,7331,7333,7349,7351,7369,7393,7411,7417,7433,7451,7457,7459,7477,7481,7487,7489,7499,7507,7517,7523,7529,7537,7541,7547,7549,7559,7561,7573,7577,7583,7589,7591,7603,7607,7621,7639,7643,7649,7669,7673,7681,7687,7691,7699,7703,7717,7723,7727,7741,7753,7757,7759,7789,7793,7817,7823,7829,7841,7853,7867,7873,7877,7879,7883,7901,7907,7919,7927,7933,7937,7949,7951,7963,7993,8009,8011,8017,8039,8053,8059,8069,8081,8087,8089,8093,8101,8111,8117,8123,8147,8161,8167,8171,8179,8191,8209,8219,8221,8231,8233,8237,8243,8263,8269,8273,8287,8291,8293,8297,8311,8317,8329,8353,8363,8369,8377,8387,8389,8419,8423,8429,8431,8443,8447,8461,8467,8501,8513,8521,8527,8537,8539,8543,8563,8573,8581,8597,8599,8609,8623,8627,8629,8641,8647,8663,8669,8677,8681,8689,8693,8699,8707,8713,8719,8731,8737,8741,8747,8753,8761,8779,8783,8803,8807,8819,8821,8831,8837,8839,8849,8861,8863,8867,8887,8893,8923,8929,8933,8941,8951,8963,8969,8971,8999,9001,9007,9011,9013,9029,9041,9043,9049,9059,9067,9091,9103,9109,9127,9133,9137,9151,9157,9161,9173,9181,9187,9199,9203,9209,9221,9227,9239,9241,9257,9277,9281,9283,9293,9311,9319,9323,9337,9341,9343,9349,9371,9377,9391,9397,9403,9413,9419,9421,9431,9433,9437,9439,9461,9463,9467,9473,9479,9491,9497,9511,9521,9533,9539,9547,9551,9587,9601,9613,9619,9623,9629,9631,9643,9649,9661,9677,9679,9689,9697,9719,9721,9733,9739,9743,9749,9767,9769,9781,9787,9791,9803,9811,9817,9829,9833,9839,9851,9857,9859,9871,9883,9887,9901,9907,9923,9929,9931,9941,9949,9967,9973};int main() {int n;while (scanf("%d", &n), n){int ans, st, en, sum;st = en = ans = sum = 0;while (1){if (sum == n) ans++;if (sum >= n) sum -= prime[st++];else{if (prime[en] <= n) sum += prime[en++];else break;}}printf("%d\n", ans);} }

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