題目描述：

給定一個數組，判斷數組內是否存在一個連續區間，使其和恰好等于給定整數k。

輸入：

輸入包含多組測試用例，每組測試用例由一個整數n(1<=n<=10000)開頭，代表數組的大小。
接下去一行為n個整數，描述這個數組，整數絕對值不大于100。
最后一行為一個整數k(大小在int范圍內)。

輸出：

對于每組測試用例，若存在這個連續區間，輸出其開始和結束的位置，s，e(s <= e)。
若存在多個符合條件的輸出，則輸出s較小的那個，若仍然存在多個，輸出e較小的那個。
若不存在，直接輸出"No"。

樣例輸入：

5 -1 2 3 -4 9 5 3 -1 2 -3 7 2 -1 1 0

樣例輸出：

2 3 No 1 2

for(int k=0;k<=10000;k++){cout<<k<<',';}
生成數組 ?拷貝進? int n[10001]={};

/** rty.cpp** Created on: 2014年12月29日* Author: judyge*/#include <iostream> #include<time.h> #include<windows.h> #include<stdio.h> using namespace std;int main() {//for(int k=0;k<=10000;k++)//{//cout<<k<<',';//}clock_t start,finish;double time;start=clock();int n[1001]={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200,201,202,203,204,205,206,207,208,209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,224,225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,240,241,242,243,244,245,246,247,248,249,250,251,252,253,254,255,256,257,258,259,260,261,262,263,264,265,266,267,268,269,270,271,272,273,274,275,276,277,278,279,280,281,282,283,284,285,286,287,288,289,290,291,292,293,294,295,296,297,298,299,300,301,302,303,304,305,306,307,308,309,310,311,312,313,314,315,316,317,318,319,320,321,322,323,324,325,326,327,328,329,330,331,332,333,334,335,336,337,338,339,340,341,342,343,344,345,346,347,348,349,350,351,352,353,354,355,356,357,358,359,360,361,362,363,364,365,366,367,368,369,370,371,372,373,374,375,376,377,378,379,380,381,382,383,384,385,386,387,388,389,390,391,392,393,394,395,396,397,398,399,400,401,402,403,404,405,406,407,408,409,410,411,412,413,414,415,416,417,418,419,420,421,422,423,424,425,426,427,428,429,430,431,432,433,434,435,436,437,438,439,440,441,442,443,444,445,446,447,448,449,450,451,452,453,454,455,456,457,458,459,460,461,462,463,464,465,466,467,468,469,470,471,472,473,474,475,476,477,478,479,480,481,482,483,484,485,486,487,488,489,490,491,492,493,494,495,496,497,498,499,500,501,502,503,504,505,506,507,508,509,510,511,512,513,514,515,516,517,518,519,520,521,522,523,524,525,526,527,528,529,530,531,532,533,534,535,536,537,538,539,540,541,542,543,544,545,546,547,548,549,550,551,552,553,554,555,556,557,558,559,560,561,562,563,564,565,566,567,568,569,570,571,572,573,574,575,576,577,578,579,580,581,582,583,584,585,586,587,588,589,590,591,592,593,594,595,596,597,598,599,600,601,602,603,604,605,606,607,608,609,610,611,612,613,614,615,616,617,618,619,620,621,622,623,624,625,626,627,628,629,630,631,632,633,634,635,636,637,638,639,640,641,642,643,644,645,646,647,648,649,650,651,652,653,654,655,656,657,658,659,660,661,662,663,664,665,666,667,668,669,670,671,672,673,674,675,676,677,678,679,680,681,682,683,684,685,686,687,688,689,690,691,692,693,694,695,696,697,698,699,700,701,702,703,704,705,706,707,708,709,710,711,712,713,714,715,716,717,718,719,720,721,722,723,724,725,726,727,728,729,730,731,732,733,734,735,736,737,738,739,740,741,742,743,744,745,746,747,748,749,750,751,752,753,754,755,756,757,758,759,760,761,762,763,764,765,766,767,768,769,770,771,772,773,774,775,776,777,778,779,780,781,782,783,784,785,786,787,788,789,790,791,792,793,794,795,796,797,798,799,800,801,802,803,804,805,806,807,808,809,810,811,812,813,814,815,816,817,818,819,820,821,822,823,824,825,826,827,828,829,830,831,832,833,834,835,836,837,838,839,840,841,842,843,844,845,846,847,848,849,850,851,852,853,854,855,856,857,858,859,860,861,862,863,864,865,866,867,868,869,870,871,872,873,874,875,876,877,878,879,880,881,882,883,884,885,886,887,888,889,890,891,892,893,894,895,896,897,898,899,900,901,902,903,904,905,906,907,908,909,910,911,912,913,914,915,916,917,918,919,920,921,922,923,924,925,926,927,928,929,930,931,932,933,934,935,936,937,938,939,940,941,942,943,944,945,946,947,948,949,950,951,952,953,954,955,956,957,958,959,960,961,962,963,964,965,966,967,968,969,970,971,972,973,974,975,976,977,978,979,980,981,982,983,984,985,986,987,988,989,990,991,992,993,994,995,996,997,998,999,1000};int k=275;int sum=0;bool flag=false;int m=sizeof(n)/sizeof(int);for(int i=0;i<=m;i++){if(flag==true){break;}sum=0;for(int j=i;j<=m;j++ ){sum=sum+n[j];if(sum==k){cout<<i<<'\n';cout<<j<<'\n';flag=true;break;}}}finish=clock();time=(double)((finish-start)/CLOCKS_PER_SEC);printf("start:%ld\t\tfinish:%ld\tfinish-start:%ld\truntime:%f\n",start,finish,finish-start,time);return 0; }

運行結果分析

2 23 start:4 finish:4 finish-start:0 runtime:0.000000

沒有超時和輸入有關.

空間換時間法

#include <stdio.h> #include <string.h> #include <queue> using namespace std; int SumList[10010]; int NumList[10010]; priority_queue<int> mask[2000010]; int n, k; int main() { while(scanf("%d", &n) != EOF) { int sum; memset(SumList, 0, sizeof(SumList)); for(int i = 1; i <= n; i++) { scanf("%d", &NumList[i]); SumList[i] = NumList[i] + SumList[i - 1]; mask[SumList[i] + 1000001].push(i); } scanf("%d", &k); bool res = false; int temp, tempIndex; for(int i = 0; i < n; i++) { tempIndex = SumList[i] + k + 1000001; if(mask[tempIndex].size()) { if(mask[tempIndex].top() < i + 1) { continue; } while(mask[tempIndex].size() && mask[tempIndex].top() >= i + 1) { temp = mask[tempIndex].top(); mask[tempIndex].pop(); } printf("%d %d\n", i + 1, temp); res = true; break; } } if(!res) { printf("No\n"); } for(int i = 1; i <= n; i++) { tempIndex = SumList[i] + 1000001; if(mask[tempIndex].size()) { mask[tempIndex].pop(); } } } return 0; }

總結

以上是生活随笔為你收集整理的九度OJ 区间问题的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。